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Praxis Core Math

Course: praxis core math   >   unit 1.

  • Algebraic properties | Lesson
  • Algebraic properties | Worked example
  • Solution procedures | Lesson
  • Solution procedures | Worked example
  • Equivalent expressions | Lesson
  • Equivalent expressions | Worked example
  • Creating expressions and equations | Lesson
  • Creating expressions and equations | Worked example

Algebraic word problems | Lesson

  • Algebraic word problems | Worked example
  • Linear equations | Lesson
  • Linear equations | Worked example
  • Quadratic equations | Lesson
  • Quadratic equations | Worked example

What are algebraic word problems?

What skills are needed.

  • Translating sentences to equations
  • Solving linear equations with one variable
  • Evaluating algebraic expressions
  • Solving problems using Venn diagrams

How do we solve algebraic word problems?

  • Define a variable.
  • Write an equation using the variable.
  • Solve the equation.
  • If the variable is not the answer to the word problem, use the variable to calculate the answer.

What's a Venn diagram?

  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
  • an exact decimal, like 0.75 ‍  
  • a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
  • (Choice A)   $ 4 ‍   A $ 4 ‍  
  • (Choice B)   $ 5 ‍   B $ 5 ‍  
  • (Choice C)   $ 9 ‍   C $ 9 ‍  
  • (Choice D)   $ 14 ‍   D $ 14 ‍  
  • (Choice E)   $ 20 ‍   E $ 20 ‍  
  • (Choice A)   10 ‍   A 10 ‍  
  • (Choice B)   12 ‍   B 12 ‍  
  • (Choice C)   24 ‍   C 24 ‍  
  • (Choice D)   30 ‍   D 30 ‍  
  • (Choice E)   32 ‍   E 32 ‍  
  • (Choice A)   4 ‍   A 4 ‍  
  • (Choice B)   10 ‍   B 10 ‍  
  • (Choice C)   14 ‍   C 14 ‍  
  • (Choice D)   18 ‍   D 18 ‍  
  • (Choice E)   22 ‍   E 22 ‍  

Things to remember

Want to join the conversation.

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Equation Word Problems Worksheets

This compilation of a meticulously drafted equation word problems worksheets is designed to get students to write and solve a variety of one-step, two-step and multi-step equations that involve integers, fractions, and decimals. These worksheets are best suited for students in grade 6 through high school. Click on the 'Free' icons to sample our handouts.

One Step Equation Word Problem Worksheets

One Step Equation Word Problem Worksheets

Read and solve this series of word problems that involve one-step equations. Apply basic operations to find the value of unknowns.

(15 Worksheets)

Two-Step Equation Word Problems: Integers

Two-Step Equation Word Problems: Integers

Interpret this set of word problems that require two-step operations to solve the equations. Each printable worksheet has five word problems ideal for 6th grade, 7th grade, and 8th grade students.

  • Download the set

Multi-Step Equation Word Problems: Integers

Multi-Step Equation Word Problems: Integers

Read each multi-step word problem in these high school pdf worksheets and set up the equation. Solve and find the value of the unknown. More than two steps are required to solve the problems.

Two-Step Equation Word Problems: Fractions and Decimals

Two-Step Equation Word Problems: Fractions and Decimals

Read each word problem and set up the two-step equation. Solve the equation and find the solution. This selection of worksheets includes both fractions and decimals.

Multi-Step Equation Word Problems: Fractions and Decimals

Multi-Step Equation Word Problems: Fractions and Decimals

Write multi-step equations that involve both fractions and decimals based on the word problems provided here. Validate your responses with our answer keys.

MCQ - Equation Word Problems

MCQ - Equation Word Problems

Pick the correct two-step equation that best matches word problems presented here. Evaluate the ability of students to solve two-step equations with this array of MCQ worksheets.

Related Worksheets

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Mathematics LibreTexts

1.20: Word Problems for Linear Equations

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  • Page ID 45640

Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable \(x .\) Twice the variable is \(2 x .\) Adding \(2 x\) to 4 gives:

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

Solution: Three times a number is \(3 x .\) We need to subtract \(3 x\) from 7. This means:\

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by \(x .8\) less than \(x\) mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

For example, 8 less than 10 is \(10-8=2\).

d) Subtract \(5 p^{2}-7 p+2\) from \(3 p^{2}+4 p\) and simplify.

Solution: We need to calculate \(3 p^{2}+4 p\) minus \(5 p^{2}-7 p+2:\)

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

e) The amount of money given by \(x\) dimes and \(y\) quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of \(10 x\) cents. Each quarter is worth 25 cents, so that this gives a total of \(25 y\) cents. Adding the two amounts gives a total of

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is \(13 .\) Find the number.

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

We solve this for \(x\). First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives \(x=\frac{18}{2}=9\).

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

We solve this as follows. First, add \(x\) :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is \(x=\frac{9}{3}=3\)

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

Dividing both sides by 4 gives the answer: \(x=3\).

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as \(x+9,\) while subtracting 7 from three times the number is written as \(3 x-7\). We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

We solve for \(x\) by adding 7 on both sides of the equation:

\[x + 16 = 3x.\nonumber\]

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

After dividing by \(2,\) we obtain the answer \(x=8\)

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by \(5 \%\). How much does the loaf of bread cost now, when its price was \(\$ 2.40\) last year?

Solution: We calculate the price increase as \(5 \% \cdot \$ 2.40 .\) We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

The new price is therefore \(\$ 2.52\).

b) To complete a job, three workers get paid at a rate of \(\$ 12\) per hour. If the total pay for the job was \(\$ 180,\) then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by \(x\). Then the total price is calculated as the price per hour \((\$ 12)\) times the number of workers times the number of hours \((3) .\) We obtain the equation

\[12 \cdot 3 \cdot x=180\nonumber\]

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

\[x=\frac{300}{5}=60\nonumber\]

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is \(4 \times 60\) feet \(=240\) feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by \(x .\) If 4 blocks weigh \(28,\) then a block weighs \(x=\frac{28}{4}=7\)

How many blocks weigh \(70 ?\) Well, we only need to find \(\frac{70}{7}=10 .\) So, the answer is \(10 .\)

Note You can solve this problem by setting up and solving the fractional equation \(\frac{28}{4}=\frac{70}{x}\). Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by \(x\). Then the length is \(2 x+3\). The perimeter is 20 in on one hand and \(2(\)length\()+2(\)width\()\) on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

f) If a circle has circumference 4in, what is its radius?

Solution: We know that \(C=2 \pi r\) where \(C\) is the circumference and \(r\) is the radius. So in this case

\[4=2 \pi r\nonumber\]

Dividing both sides by \(2 \pi\) gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let \(x\) equal the side of the triangle. Then the perimeter is, on the one hand, \(60,\) and on other hand \(3 x .\) So \(3 x=60\) and dividing both sides of the equation by 3 gives \(x=20\) meters.

h) If a gardener has \(\$ 600\) to spend on a fence which costs \(\$ 10\) per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is \(P=2 L+2 W\). Let \(x\) be the width of the rectangle. Then the length is \(2 x .\) The perimeter is \(P=2(2 x)+2 x=6 x\). The largest perimeter is \(\$ 600 /(\$ 10 / f t)=60\) ft. So \(60=6 x\) and dividing both sides by 6 gives \(x=60 / 6=10\). So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let \(b\) be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is \(\frac{1}{2}(3.2+b) \cdot 4=\) \(6.4+2 b .\) So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

  • Read the whole thing first
  • Do a sketch if possible
  • Assign letters for the values
  • Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

  • Let S = dollars Sam has
  • Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

  • Let D = number of dogs
  • Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

  • Use w for width of rectangle: w = 12m
  • Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

tennis

Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

  • Use S for how many games Sam played
  • Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

table

Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

  • Use a for Alex's work rate
  • Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

track

Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

  • The number of "5 hour" days: d
  • The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

Example: A circle has an area of 12 mm 2 , what is its radius?

  • Use A for Area: A = 12 mm 2
  • Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

  • Use V for Volume
  • Use A for Area
  • Use s for side length of cube
  • Volume of a cube: V = s 3
  • Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

pizza

Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

  • Joel's normal rate of pay: $N per hour
  • Joel works for 40 hours at $N per hour = $40N
  • When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
  • Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
  • And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

  • Present Value PV = $2,000
  • Interest Rate (as a decimal): r = 0.11
  • Number of Periods: n = 3
  • Future Value (the value we want): FV

We are being asked for the Future Value: FV

Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

  • Present Value PV = $1,000
  • Interest Rate (the value we want): r
  • Number of Periods: n = 9
  • Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

  • Number of boys now: b
  • Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

  • We could try, say, n=10: 10(12) = 120 NO (too small)
  • Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side. Your client has 56 square meters of beautiful marble tiles to cover the whole area. What should the length of the room be?

Let's first make a sketch so we get things right!:

  • the length of the room: L
  • the width of the room: W
  • the total Area including veranda: A
  • the width of the room is half its length: W = ½L
  • the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

This is a quadratic equation , there are many ways to solve it, this time let's use factoring :

And so L = 8 or −14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?

Equation Word Problems — Examples & Practice - Expii

Equation word problems — examples & practice, explanations (3).

solving equations word problem examples

Writing Equations for Word Problems

Here is a refresher on translating words to math in case you need it for this example!

solving equations word problem examples

Image by Caroline Kulczycky

Related Lessons

solving equations word problem examples

(Video) Writing Equations from Word Problems

by Katie Tatum

solving equations word problem examples

Here's a video by Katie Tatum that shows you how to create equations from a word problem.

This is a very important and useful skill to learn how to do. The trick is to locate key words that signal you to what needs to be done. For example, "fewer" means subtraction . If there's a vague term like "there's a number" then that means you should use a variable like n. We can use all the information to write an expression. That expression turns the words into algebra. Then there will usually be some sort of calculation to do.

Note to those who aren't good at word problems: Personally, I've never been good at word problems, and I think a lot of people feel the same. Now that I'm older and farther into a math career I look back and find the problems—that I used to find difficult—super easy. So, don't fret. Just practice and try your best and one day, without realizing it, you'll come to understand them.

Test Yourself: Michael has twice as many baseball cards this years than he did last year. He gives a third of these to his friend Jason. Now, Michael has 40 cards. How many did he have a year ago? This one is a little tricky.

Translate spoken language into math

Let's solve some word problems in algebra. You're first going to have to first translate the problem into an equation. In other words, we set up the equation. Finally, solve the equation].

Word problems can be as simple as "find a number that's three less than twelve" or as complicated as...well, they can get really complicated.

There are lots of patterns to how words translate into numbers and operations, but it'll be tough to learn if you just try to memorize them. Instead, we're going to work through the "less than twelve" example above to practice asking questions that will help you translate.

The questions we'll ask here are:

  • What are the numbers in the problem?
  • What are the operations in the problem?
  • How do the numbers relate to each other?

Let's start off with the easy question. What are the numbers in the problem? "Find a number that's three less than twelve".

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Algebra Topics  - Introduction to Word Problems

Algebra topics  -, introduction to word problems, algebra topics introduction to word problems.

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Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

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What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you've ever taken a math class, you've probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you're supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 - 4 = 8 , so you know Johnny has 8 apples left.

Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

  • Read through the problem carefully, and figure out what it's about.
  • Represent unknown numbers with variables.
  • Translate the rest of the problem into a mathematical expression.
  • Solve the problem.
  • Check your work.

We'll work through an algebra word problem using these steps. Here's a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let's go through it step by step.

Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you're reading, consider:

  • What question is the problem asking?
  • What information do you already have?

Let's take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There's only one question here. We're trying to find out how many miles Jada drove . Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

  • The van cost $30 per day.
  • In addition to paying a daily charge, Jada paid $0.50 per mile.
  • Jada had the van for 2 days.
  • The total cost was $360 .

Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables . (To learn more about variables, see our lesson on reading algebraic expressions .) You can use a variable in the place of any amount you don't know. Looking at our problem, do you see a quantity we should represent with a variable? It's often the number we're trying to find out.

Since we're trying to find the total number of miles Jada drove, we'll represent that amount with a variable—at least until we know it. We'll use the variable m for miles . Of course, we could use any variable, but m should be easy to remember.

Step 3: Translate the rest of the problem.

Let's take another look at the problem, with the facts we'll use to solve it highlighted.

The rate to rent a small moving van is $30 per day , plus $0.50 per mile . Jada rented a van to drive to her new home. It took 2 days , and the van cost $360 . How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It's $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360 . The shorter version will be easier to translate into a mathematical expression.

Let's start by translating $30 per day . To calculate the cost of something that costs a certain amount per day, you'd multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅ days, or 30 times the number of days . (Not sure why you'd translate it this way? Check out our lesson on writing algebraic expressions .)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50 , $.50 per mile became $.50 ⋅ mile , and is became = .

Next, we'll add in the numbers and variables we already know. We already know the number of days Jada drove, 2 , so we can replace that. We've also already said we'll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that's left to do is solve it.

Step 4: Solve the problem.

This problem will take a few steps to solve. (If you're not sure how to do the math in this section, you might want to review our lesson on simplifying expressions .) First, let's simplify the expression as much as possible. We can multiply 30 and 2, so let's go ahead and do that. We can also write .5 ⋅ m as 0.5 m .

60 + .5m = 360

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we'll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides .

The only thing left to get rid of is .5 . Since it's being multiplied with m , we'll do the reverse and divide both sides of the equation with it.

.5 m / .5 is m and 300 / 0.50 is 600 , so m = 600 . In other words, the answer to our problem is 600 —we now know Jada drove 600 miles.

Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got— 600 —and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada's distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let's take another look at the problem.

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We're done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Let's practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Try completing this problem on your own. When you're done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Here's another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Problem 1 Answer

Here's Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Answer: $29

Let's solve this problem step by step. We'll solve it the same way we solved the problem on page 1.

Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it . Let's look at the problem again. The question is right there in plain sight:

So is the information we'll need to answer the question:

  • A single ticket costs $8 .
  • The family pass costs $25 more than half the price of the single ticket.

Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass . We'll represent it with the variable f .

Step 3: Translate the rest of the problem

Let's look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8 . A family pass costs $25 more than half that . How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25 . To turn this into a problem we can solve, we'll have to translate it into math. Here's how:

  • First, replace the cost of a family pass with our variable f .

f equals half of $8 plus $25

  • Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

  • Finally, translate the rest of the problem. Half of can be written as 1/2 times , or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

  • f is already alone on the left side of the equation, so all we have to do is calculate the right side.
  • First, multiply 1/2 by 8 . 1/2 ⋅ 8 is 4 .
  • Next, add 4 and 25. 4 + 25 equals 29 .

That's it! f is equal to 29. In other words, the cost of a family pass is $29 .

Step 5: Check your work

Finally, let's check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let's look at the original problem again.

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

  • We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 - 25

  • Let's work on the right side first. 29 - 25 is 4 .
  • To find the value of s , we have to get it alone on the left side of the equation. This means getting rid of 1/2 . To do this, we'll multiply each side by the inverse of 1/2: 2 .

According to our math, s = 8 . In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that's correct!

So now we're sure about the answer to our problem: The cost of a family pass is $29 .

Problem 2 Answer

Here's Problem 2:

Answer: $70

Let's go through this problem one step at a time.

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it . What's the question here?

To solve the problem, you'll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

  • The amount Flor donated is three times as much the amount Mo donated
  • Flor and Mo's donations add up to $280 total

The unknown number we're trying to identify in this problem is Mo's donation . We'll represent it with the variable m .

Here's the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo . Between the two of them, they donated $280 . How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo's donation plus Flor's donation equals $280

Because we know that Flor's donation is three times as much as Mo's donation, we could go even further and say:

Mo's donation plus three times Mo's donation equals $280

We can translate this into a math problem in only a few steps. Here's how:

  • Because we've already said we'll represent the amount of Mo's donation with the variable m , let's start by replacing Mo's donation with m .

m plus three times m equals $280

  • Next, we can put in mathematical operators in place of certain words. We'll also take out the dollar sign.

m + three times m = 280

  • Finally, let's write three times mathematically. Three times m can also be written as 3 ⋅ m , or just 3 m .

m + 3m = 280

It will only take a few steps to solve this problem.

  • To get the correct answer, we'll have to get m alone on one side of the equation.
  • To start, let's add m and 3 m . That's 4 m .
  • We can get rid of the 4 next to the m by dividing both sides by 4. 4 m / 4 is m , and 280 / 4 is 70 .

We've got our answer: m = 70 . In other words, Mo donated $70 .

The answer to our problem is $70 , but we should check just to be sure. Let's look at our problem again.

If our answer is correct, $70 and three times $70 should add up to $280 .

  • We can write our new equation like this:

70 + 3 ⋅ 70 = 280

  • The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

  • The last step is to add 70 and 210. 70 plus 210 equals 280 .

280 is the combined cost of the tickets in our original problem. Our answer is correct : Mo gave $70 to charity.

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120 Math Word Problems To Challenge Students Grades 1 to 8

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Written by Marcus Guido

Hey teachers! 👋

Use Prodigy to spark a love for math in your students – including when solving word problems!

  • Teaching Tools
  • Subtraction
  • Multiplication
  • Mixed operations
  • Ordering and number sense
  • Comparing and sequencing
  • Physical measurement
  • Ratios and percentages
  • Probability and data relationships

You sit at your desk, ready to put a math quiz, test or activity together. The questions flow onto the document until you hit a section for word problems.

A jolt of creativity would help. But it doesn’t come.

Whether you’re a 3rd grade teacher or an 8th grade teacher preparing students for high school, translating math concepts into real world examples can certainly be a challenge.

This resource is your jolt of creativity. It provides examples and templates of math word problems for 1st to 8th grade classes.

There are 120 examples in total.

The list of examples is supplemented by tips to create engaging and challenging math word problems.

120 Math word problems, categorized by skill

Addition word problems.

A teacher is teaching three students with a whiteboard happily.

Best for: 1st grade, 2nd grade

1. Adding to 10: Ariel was playing basketball. 1 of her shots went in the hoop. 2 of her shots did not go in the hoop. How many shots were there in total?

2. Adding to 20: Adrianna has 10 pieces of gum to share with her friends. There wasn’t enough gum for all her friends, so she went to the store to get 3 more pieces of gum. How many pieces of gum does Adrianna have now?

3. Adding to 100: Adrianna has 10 pieces of gum to share with her friends. There wasn’t enough gum for all her friends, so she went to the store and got 70 pieces of strawberry gum and 10 pieces of bubble gum. How many pieces of gum does Adrianna have now?

4. Adding Slightly over 100: The restaurant has 175 normal chairs and 20 chairs for babies. How many chairs does the restaurant have in total?

5. Adding to 1,000: How many cookies did you sell if you sold 320 chocolate cookies and 270 vanilla cookies?

6. Adding to and over 10,000: The hobby store normally sells 10,576 trading cards per month. In June, the hobby store sold 15,498 more trading cards than normal. In total, how many trading cards did the hobby store sell in June?

7. Adding 3 Numbers: Billy had 2 books at home. He went to the library to take out 2 more books. He then bought 1 book. How many books does Billy have now?

8. Adding 3 Numbers to and over 100: Ashley bought a big bag of candy. The bag had 102 blue candies, 100 red candies and 94 green candies. How many candies were there in total?

Subtraction word problems

Best for: 1st grade, second grade

9. Subtracting to 10: There were 3 pizzas in total at the pizza shop. A customer bought 1 pizza. How many pizzas are left?

10. Subtracting to 20: Your friend said she had 11 stickers. When you helped her clean her desk, she only had a total of 10 stickers. How many stickers are missing?

11. Subtracting to 100: Adrianna has 100 pieces of gum to share with her friends. When she went to the park, she shared 10 pieces of strawberry gum. When she left the park, Adrianna shared another 10 pieces of bubble gum. How many pieces of gum does Adrianna have now?

Five middle school students sitting at a row of desks playing Prodigy Math on tablets.

Practice math word problems with Prodigy Math

Join millions of teachers using Prodigy to make learning fun and differentiate instruction as they answer in-game questions, including math word problems from 1st to 8th grade!

12. Subtracting Slightly over 100: Your team scored a total of 123 points. 67 points were scored in the first half. How many were scored in the second half?

13. Subtracting to 1,000: Nathan has a big ant farm. He decided to sell some of his ants. He started with 965 ants. He sold 213. How many ants does he have now?

14. Subtracting to and over 10,000: The hobby store normally sells 10,576 trading cards per month. In July, the hobby store sold a total of 20,777 trading cards. How many more trading cards did the hobby store sell in July compared with a normal month?

15. Subtracting 3 Numbers: Charlene had a pack of 35 pencil crayons. She gave 6 to her friend Theresa. She gave 3 to her friend Mandy. How many pencil crayons does Charlene have left?

16. Subtracting 3 Numbers to and over 100: Ashley bought a big bag of candy to share with her friends. In total, there were 296 candies. She gave 105 candies to Marissa. She also gave 86 candies to Kayla. How many candies were left?

Multiplication word problems

A hand holding a pen is doing calculation on a pice of papper

Best for: 2nd grade, 3rd grade

17. Multiplying 1-Digit Integers: Adrianna needs to cut a pan of brownies into pieces. She cuts 6 even columns and 3 even rows into the pan. How many brownies does she have?

18. Multiplying 2-Digit Integers: A movie theatre has 25 rows of seats with 20 seats in each row. How many seats are there in total?

19. Multiplying Integers Ending with 0: A clothing company has 4 different kinds of sweatshirts. Each year, the company makes 60,000 of each kind of sweatshirt. How many sweatshirts does the company make each year?

20. Multiplying 3 Integers: A bricklayer stacks bricks in 2 rows, with 10 bricks in each row. On top of each row, there is a stack of 6 bricks. How many bricks are there in total?

21. Multiplying 4 Integers: Cayley earns $5 an hour by delivering newspapers. She delivers newspapers 3 days each week, for 4 hours at a time. After delivering newspapers for 8 weeks, how much money will Cayley earn?

Division word problems

Best for: 3rd grade, 4th grade, 5th grade

22. Dividing 1-Digit Integers: If you have 4 pieces of candy split evenly into 2 bags, how many pieces of candy are in each bag?

23. Dividing 2-Digit Integers: If you have 80 tickets for the fair and each ride costs 5 tickets, how many rides can you go on?

24. Dividing Numbers Ending with 0: The school has $20,000 to buy new computer equipment. If each piece of equipment costs $50, how many pieces can the school buy in total?

25. Dividing 3 Integers: Melissa buys 2 packs of tennis balls for $12 in total. All together, there are 6 tennis balls. How much does 1 pack of tennis balls cost? How much does 1 tennis ball cost?

26. Interpreting Remainders: An Italian restaurant receives a shipment of 86 veal cutlets. If it takes 3 cutlets to make a dish, how many cutlets will the restaurant have left over after making as many dishes as possible?

Mixed operations word problems

A female teacher is instructing student math on a blackboard

27. Mixing Addition and Subtraction: There are 235 books in a library. On Monday, 123 books are taken out. On Tuesday, 56 books are brought back. How many books are there now?

28. Mixing Multiplication and Division: There is a group of 10 people who are ordering pizza. If each person gets 2 slices and each pizza has 4 slices, how many pizzas should they order?

29. Mixing Multiplication, Addition and Subtraction: Lana has 2 bags with 2 marbles in each bag. Markus has 2 bags with 3 marbles in each bag. How many more marbles does Markus have?

30. Mixing Division, Addition and Subtraction: Lana has 3 bags with the same amount of marbles in them, totaling 12 marbles. Markus has 3 bags with the same amount of marbles in them, totaling 18 marbles. How many more marbles does Markus have in each bag?

Ordering and number sense word problems

31. Counting to Preview Multiplication: There are 2 chalkboards in your classroom. If each chalkboard needs 2 pieces of chalk, how many pieces do you need in total?

32. Counting to Preview Division: There are 3 chalkboards in your classroom. Each chalkboard has 2 pieces of chalk. This means there are 6 pieces of chalk in total. If you take 1 piece of chalk away from each chalkboard, how many will there be in total?

33. Composing Numbers: What number is 6 tens and 10 ones?

34. Guessing Numbers: I have a 7 in the tens place. I have an even number in the ones place. I am lower than 74. What number am I?

35. Finding the Order: In the hockey game, Mitchell scored more points than William but fewer points than Auston. Who scored the most points? Who scored the fewest points?

Fractions word problems

A student is drawing on a notebook, holding a pencil.

Best for: 3rd grade, 4th grade, 5th grade, 6th grade

36. Finding Fractions of a Group: Julia went to 10 houses on her street for Halloween. 5 of the houses gave her a chocolate bar. What fraction of houses on Julia’s street gave her a chocolate bar?

37. Finding Unit Fractions: Heather is painting a portrait of her best friend, Lisa. To make it easier, she divides the portrait into 6 equal parts. What fraction represents each part of the portrait?

38. Adding Fractions with Like Denominators: Noah walks ⅓ of a kilometre to school each day. He also walks ⅓ of a kilometre to get home after school. How many kilometres does he walk in total?

39. Subtracting Fractions with Like Denominators: Last week, Whitney counted the number of juice boxes she had for school lunches. She had ⅗ of a case. This week, it’s down to ⅕ of a case. How much of the case did Whitney drink?

40. Adding Whole Numbers and Fractions with Like Denominators: At lunchtime, an ice cream parlor served 6 ¼ scoops of chocolate ice cream, 5 ¾ scoops of vanilla and 2 ¾ scoops of strawberry. How many scoops of ice cream did the parlor serve in total?

41. Subtracting Whole Numbers and Fractions with Like Denominators: For a party, Jaime had 5 ⅓ bottles of cola for her friends to drink. She drank ⅓ of a bottle herself. Her friends drank 3 ⅓. How many bottles of cola does Jaime have left?

42. Adding Fractions with Unlike Denominators: Kevin completed ½ of an assignment at school. When he was home that evening, he completed ⅚ of another assignment. How many assignments did Kevin complete?

43. Subtracting Fractions with Unlike Denominators: Packing school lunches for her kids, Patty used ⅞ of a package of ham. She also used ½ of a package of turkey. How much more ham than turkey did Patty use?

44. Multiplying Fractions: During gym class on Wednesday, the students ran for ¼ of a kilometre. On Thursday, they ran ½ as many kilometres as on Wednesday. How many kilometres did the students run on Thursday? Write your answer as a fraction.

45. Dividing Fractions: A clothing manufacturer uses ⅕ of a bottle of colour dye to make one pair of pants. The manufacturer used ⅘ of a bottle yesterday. How many pairs of pants did the manufacturer make?

46. Multiplying Fractions with Whole Numbers: Mark drank ⅚ of a carton of milk this week. Frank drank 7 times more milk than Mark. How many cartons of milk did Frank drink? Write your answer as a fraction, or as a whole or mixed number.

Decimals word problems

Best for: 4th grade, 5th grade

47. Adding Decimals: You have 2.6 grams of yogurt in your bowl and you add another spoonful of 1.3 grams. How much yogurt do you have in total?

48. Subtracting Decimals: Gemma had 25.75 grams of frosting to make a cake. She decided to use only 15.5 grams of the frosting. How much frosting does Gemma have left?

49. Multiplying Decimals with Whole Numbers: Marshall walks a total of 0.9 kilometres to and from school each day. After 4 days, how many kilometres will he have walked?

50. Dividing Decimals by Whole Numbers: To make the Leaning Tower of Pisa from spaghetti, Mrs. Robinson bought 2.5 kilograms of spaghetti. Her students were able to make 10 leaning towers in total. How many kilograms of spaghetti does it take to make 1 leaning tower?

51. Mixing Addition and Subtraction of Decimals: Rocco has 1.5 litres of orange soda and 2.25 litres of grape soda in his fridge. Antonio has 1.15 litres of orange soda and 0.62 litres of grape soda. How much more soda does Rocco have than Angelo?

52. Mixing Multiplication and Division of Decimals: 4 days a week, Laura practices martial arts for 1.5 hours. Considering a week is 7 days, what is her average practice time per day each week?

Comparing and sequencing word problems

Four students are sitting together and discussing math questions

Best for: Kindergarten, 1st grade, 2nd grade

53. Comparing 1-Digit Integers: You have 3 apples and your friend has 5 apples. Who has more?

54. Comparing 2-Digit Integers: You have 50 candies and your friend has 75 candies. Who has more?

55. Comparing Different Variables: There are 5 basketballs on the playground. There are 7 footballs on the playground. Are there more basketballs or footballs?

56. Sequencing 1-Digit Integers: Erik has 0 stickers. Every day he gets 1 more sticker. How many days until he gets 3 stickers?

57. Skip-Counting by Odd Numbers: Natalie began at 5. She skip-counted by fives. Could she have said the number 20?

58. Skip-Counting by Even Numbers: Natasha began at 0. She skip-counted by eights. Could she have said the number 36?

59. Sequencing 2-Digit Numbers: Each month, Jeremy adds the same number of cards to his baseball card collection. In January, he had 36. 48 in February. 60 in March. How many baseball cards will Jeremy have in April?

Time word problems

66. Converting Hours into Minutes: Jeremy helped his mom for 1 hour. For how many minutes was he helping her?

69. Adding Time: If you wake up at 7:00 a.m. and it takes you 1 hour and 30 minutes to get ready and walk to school, at what time will you get to school?

70. Subtracting Time: If a train departs at 2:00 p.m. and arrives at 4:00 p.m., how long were passengers on the train for?

71. Finding Start and End Times: Rebecca left her dad’s store to go home at twenty to seven in the evening. Forty minutes later, she was home. What time was it when she arrived home?

Money word problems

Best for: 1st grade, 2nd grade, 3rd grade, 4th grade, 5th grade

60. Adding Money: Thomas and Matthew are saving up money to buy a video game together. Thomas has saved $30. Matthew has saved $35. How much money have they saved up together in total?

61. Subtracting Money: Thomas has $80 saved up. He uses his money to buy a video game. The video game costs $67. How much money does he have left?

62. Multiplying Money: Tim gets $5 for delivering the paper. How much money will he have after delivering the paper 3 times?

63. Dividing Money: Robert spent $184.59 to buy 3 hockey sticks. If each hockey stick was the same price, how much did 1 cost?

64. Adding Money with Decimals: You went to the store and bought gum for $1.25 and a sucker for $0.50. How much was your total?

65. Subtracting Money with Decimals: You went to the store with $5.50. You bought gum for $1.25, a chocolate bar for $1.15 and a sucker for $0.50. How much money do you have left?

67. Applying Proportional Relationships to Money: Jakob wants to invite 20 friends to his birthday, which will cost his parents $250. If he decides to invite 15 friends instead, how much money will it cost his parents? Assume the relationship is directly proportional.

68. Applying Percentages to Money: Retta put $100.00 in a bank account that gains 20% interest annually. How much interest will be accumulated in 1 year? And if she makes no withdrawals, how much money will be in the account after 1 year?

Physical measurement word problems

A girl is doing math practice

Best for: 1st grade, 2nd grade, 3rd grade, 4th grade

72. Comparing Measurements: Cassandra’s ruler is 22 centimetres long. April’s ruler is 30 centimetres long. How many centimetres longer is April’s ruler?

73. Contextualizing Measurements: Picture a school bus. Which unit of measurement would best describe the length of the bus? Centimetres, metres or kilometres?

74. Adding Measurements: Micha’s dad wants to try to save money on gas, so he has been tracking how much he uses. Last year, Micha’s dad used 100 litres of gas. This year, her dad used 90 litres of gas. How much gas did he use in total for the two years?

75. Subtracting Measurements: Micha’s dad wants to try to save money on gas, so he has been tracking how much he uses. Over the past two years, Micha’s dad used 200 litres of gas. This year, he used 100 litres of gas. How much gas did he use last year?

A tablet showing an example of Prodigy Math's battle gameplay.

76. Multiplying Volume and Mass: Kiera wants to make sure she has strong bones, so she drinks 2 litres of milk every week. After 3 weeks, how many litres of milk will Kiera drink?

77. Dividing Volume and Mass: Lillian is doing some gardening, so she bought 1 kilogram of soil. She wants to spread the soil evenly between her 2 plants. How much will each plant get?

78. Converting Mass: Inger goes to the grocery store and buys 3 squashes that each weigh 500 grams. How many kilograms of squash did Inger buy?

79. Converting Volume: Shad has a lemonade stand and sold 20 cups of lemonade. Each cup was 500 millilitres. How many litres did Shad sell in total?

80. Converting Length: Stacy and Milda are comparing their heights. Stacy is 1.5 meters tall. Milda is 10 centimetres taller than Stacy. What is Milda’s height in centimetres?

81. Understanding Distance and Direction: A bus leaves the school to take students on a field trip. The bus travels 10 kilometres south, 10 kilometres west, another 5 kilometres south and 15 kilometres north. To return to the school, in which direction does the bus have to travel? How many kilometres must it travel in that direction?

Ratios and percentages word problems

Best for: 4th grade, 5th grade, 6th grade

82. Finding a Missing Number: The ratio of Jenny’s trophies to Meredith’s trophies is 7:4. Jenny has 28 trophies. How many does Meredith have?

83. Finding Missing Numbers: The ratio of Jenny’s trophies to Meredith’s trophies is 7:4. The difference between the numbers is 12. What are the numbers?

84. Comparing Ratios: The school’s junior band has 10 saxophone players and 20 trumpet players. The school’s senior band has 18 saxophone players and 29 trumpet players. Which band has the higher ratio of trumpet to saxophone players?

85. Determining Percentages: Mary surveyed students in her school to find out what their favourite sports were. Out of 1,200 students, 455 said hockey was their favourite sport. What percentage of students said hockey was their favourite sport?

86. Determining Percent of Change: A decade ago, Oakville’s population was 67,624 people. Now, it is 190% larger. What is Oakville’s current population?

87. Determining Percents of Numbers: At the ice skate rental stand, 60% of 120 skates are for boys. If the rest of the skates are for girls, how many are there?

88. Calculating Averages: For 4 weeks, William volunteered as a helper for swimming classes. The first week, he volunteered for 8 hours. He volunteered for 12 hours in the second week, and another 12 hours in the third week. The fourth week, he volunteered for 9 hours. For how many hours did he volunteer per week, on average?

Probability and data relationships word problems

Two students are calculating on a whiteboard

Best for: 4th grade, 5th grade, 6th grade, 7th grade

89. Understanding the Premise of Probability: John wants to know his class’s favourite TV show, so he surveys all of the boys. Will the sample be representative or biased?

90. Understanding Tangible Probability: The faces on a fair number die are labelled 1, 2, 3, 4, 5 and 6. You roll the die 12 times. How many times should you expect to roll a 1?

91. Exploring Complementary Events: The numbers 1 to 50 are in a hat. If the probability of drawing an even number is 25/50, what is the probability of NOT drawing an even number? Express this probability as a fraction.

92. Exploring Experimental Probability: A pizza shop has recently sold 15 pizzas. 5 of those pizzas were pepperoni. Answering with a fraction, what is the experimental probability that he next pizza will be pepperoni?

93. Introducing Data Relationships: Maurita and Felice each take 4 tests. Here are the results of Maurita’s 4 tests: 4, 4, 4, 4. Here are the results for 3 of Felice’s 4 tests: 3, 3, 3. If Maurita’s mean for the 4 tests is 1 point higher than Felice’s, what’s the score of Felice’s 4th test?

94. Introducing Proportional Relationships: Store A is selling 7 pounds of bananas for $7.00. Store B is selling 3 pounds of bananas for $6.00. Which store has the better deal?

95. Writing Equations for Proportional Relationships: Lionel loves soccer, but has trouble motivating himself to practice. So, he incentivizes himself through video games. There is a proportional relationship between the amount of drills Lionel completes, in x , and for how many hours he plays video games, in y . When Lionel completes 10 drills, he plays video games for 30 minutes. Write the equation for the relationship between x and y .

Geometry word problems

Best for: 4th grade, 5th grade, 6th grade, 7th grade, 8th grade

96. Introducing Perimeter:  The theatre has 4 chairs in a row. There are 5 rows. Using rows as your unit of measurement, what is the perimeter?

97. Introducing Area: The theatre has 4 chairs in a row. There are 5 rows. How many chairs are there in total?

98. Introducing Volume: Aaron wants to know how much candy his container can hold. The container is 20 centimetres tall, 10 centimetres long and 10 centimetres wide. What is the container’s volume?

99. Understanding 2D Shapes: Kevin draws a shape with 4 equal sides. What shape did he draw?

100. Finding the Perimeter of 2D Shapes: Mitchell wrote his homework questions on a piece of square paper. Each side of the paper is 8 centimetres. What is the perimeter?

101. Determining the Area of 2D Shapes: A single trading card is 9 centimetres long by 6 centimetres wide. What is its area?

102. Understanding 3D Shapes: Martha draws a shape that has 6 square faces. What shape did she draw?

103. Determining the Surface Area of 3D Shapes: What is the surface area of a cube that has a width of 2cm, height of 2 cm and length of 2 cm?

104. Determining the Volume of 3D Shapes: Aaron’s candy container is 20 centimetres tall, 10 centimetres long and 10 centimetres wide. Bruce’s container is 25 centimetres tall, 9 centimetres long and 9 centimetres wide. Find the volume of each container. Based on volume, whose container can hold more candy?

105. Identifying Right-Angled Triangles: A triangle has the following side lengths: 3 cm, 4 cm and 5 cm. Is this triangle a right-angled triangle?

106. Identifying Equilateral Triangles: A triangle has the following side lengths: 4 cm, 4 cm and 4 cm. What kind of triangle is it?

107. Identifying Isosceles Triangles: A triangle has the following side lengths: 4 cm, 5 cm and 5 cm. What kind of triangle is it?

108. Identifying Scalene Triangles: A triangle has the following side lengths: 4 cm, 5 cm and 6 cm. What kind of triangle is it?

109. Finding the Perimeter of Triangles: Luigi built a tent in the shape of an equilateral triangle. The perimeter is 21 metres. What is the length of each of the tent’s sides?

110. Determining the Area of Triangles: What is the area of a triangle with a base of 2 units and a height of 3 units?

111. Applying Pythagorean Theorem: A right triangle has one non-hypotenuse side length of 3 inches and the hypotenuse measures 5 inches. What is the length of the other non-hypotenuse side?

112. Finding a Circle’s Diameter: Jasmin bought a new round backpack. Its area is 370 square centimetres. What is the round backpack’s diameter?

113. Finding a Circle's Area: Captain America’s circular shield has a diameter of 76.2 centimetres. What is the area of his shield?

114. Finding a Circle’s Radius: Skylar lives on a farm, where his dad keeps a circular corn maze. The corn maze has a diameter of 2 kilometres. What is the maze’s radius?

Variables word problems

A hand is calculating math problem on a blacboard

Best for: 6th grade, 7th grade, 8th grade

115. Identifying Independent and Dependent Variables: Victoria is baking muffins for her class. The number of muffins she makes is based on how many classmates she has. For this equation, m is the number of muffins and c is the number of classmates. Which variable is independent and which variable is dependent?

116. Writing Variable Expressions for Addition: Last soccer season, Trish scored g goals. Alexa scored 4 more goals than Trish. Write an expression that shows how many goals Alexa scored.

117. Writing Variable Expressions for Subtraction: Elizabeth eats a healthy, balanced breakfast b times a week. Madison sometimes skips breakfast. In total, Madison eats 3 fewer breakfasts a week than Elizabeth. Write an expression that shows how many times a week Madison eats breakfast.

118. Writing Variable Expressions for Multiplication: Last hockey season, Jack scored g goals. Patrik scored twice as many goals than Jack. Write an expression that shows how many goals Patrik scored.

119. Writing Variable Expressions for Division: Amanda has c chocolate bars. She wants to distribute the chocolate bars evenly among 3 friends. Write an expression that shows how many chocolate bars 1 of her friends will receive.

120. Solving Two-Variable Equations: This equation shows how the amount Lucas earns from his after-school job depends on how many hours he works: e = 12h . The variable h represents how many hours he works. The variable e represents how much money he earns. How much money will Lucas earn after working for 6 hours?

How to easily make your own math word problems & word problems worksheets

Two teachers are discussing math with a pen and a notebook

Armed with 120 examples to spark ideas, making your own math word problems can engage your students and ensure alignment with lessons. Do:

  • Link to Student Interests:  By framing your word problems with student interests, you’ll likely grab attention. For example, if most of your class loves American football, a measurement problem could involve the throwing distance of a famous quarterback.
  • Make Questions Topical:  Writing a word problem that reflects current events or issues can engage students by giving them a clear, tangible way to apply their knowledge.
  • Include Student Names:  Naming a question’s characters after your students is an easy way make subject matter relatable, helping them work through the problem.
  • Be Explicit:  Repeating keywords distills the question, helping students focus on the core problem.
  • Test Reading Comprehension:  Flowery word choice and long sentences can hide a question’s key elements. Instead, use concise phrasing and grade-level vocabulary.
  • Focus on Similar Interests:  Framing too many questions with related interests -- such as football and basketball -- can alienate or disengage some students.
  • Feature Red Herrings:  Including unnecessary information introduces another problem-solving element, overwhelming many elementary students.

A key to differentiated instruction , word problems that students can relate to and contextualize will capture interest more than generic and abstract ones.

Final thoughts about math word problems

You’ll likely get the most out of this resource by using the problems as templates, slightly modifying them by applying the above tips. In doing so, they’ll be more relevant to -- and engaging for -- your students.

Regardless, having 120 curriculum-aligned math word problems at your fingertips should help you deliver skill-building challenges and thought-provoking assessments.

The result?

A greater understanding of how your students process content and demonstrate understanding, informing your ongoing teaching approach.

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7.3 Simple Algebraic Equations and Word Problems

An algebraic equation is a mathematical sentence expressing equality between two algebraic expressions (or an algebraic expression and a number).

When two expressions are joined by an equal (=) sign, it indicates that the expression to the left of the equal sign is identical in value to the expression to the right of the equal sign.

For example, when two algebraic expressions, such as [latex]5x + 7[/latex] and [latex]x + 19[/latex], are equal, the two expressions are joined by an equal (=) sign and the equation is written as:

[latex]5x + 7 = x + 19[/latex]

‘Left side’ (LS) = ‘Right side’ (RS)

The solution to the equation is the value of the variable that makes the left side (LS) evaluate to the same number as the right side (RS).

Note: You need an equation to solve for an unknown variable – you cannot solve for a variable in an algebraic expression that is not part of an equation.

  • If you have an expression , it needs to be simplified .
  • If you have an equation , it needs to be solved .

In algebra, there are a variety of equations. In this section, we will learn one equation category, linear equations with one variable .

Examples of linear equations with one variable are:

[latex]2x = 8[/latex],   [latex]3x + 5 = 14[/latex],   [latex]5x + 7 = x + 19[/latex]

An equation is either true or false depending on the value of the variable.

For example, consider the equation [latex]2x = 8[/latex]:

  • If [latex]x = 4[/latex],   LS = 2(4) = 8, RS = 8; therefore, the equation is true.
  • If [latex]x = 3[/latex],   LS = 2(3) = 6, RS = 8; therefore, the equation is false.

Equations may be classified into the following three types:

  • Conditional equation: these equations are only true when the variable has a specific value. For example, [latex]2x = 8[/latex] is a conditional equation, true if and only if [latex]x = 4[/latex].
  • Identity: these equations are true for any value for the variable. For example, [latex]2x + 10 = 2(x + 5)[/latex] is an identity, true for any value of [latex]x[/latex].
  • Contradiction: these equations are not true for any value of the variable. For example, [latex]x + 5 = x + 4[/latex] is a contradiction, not true for any value of [latex]x[/latex].

Equivalent Equations

Equations with the same solutions are called equivalent equations .

For example, [latex]2x + 5 = 9[/latex] and [latex]2x = 4[/latex] are equivalent equations because the solution [latex]x = 2[/latex] satisfies each equation.

Similarly, [latex]3x - 4 = 5[/latex], [latex]2x = x + 3[/latex], and [latex]x + 1 = 4[/latex] are equivalent equations because the solution [latex]x = 3[/latex] satisfies each equation.

Properties of Equality

If [latex]a = b[/latex], then,

These properties are used to solve equations.

Equations with Fractional Coefficients

If an equation contains fractional coefficients, then the fractional coefficients can be changed to whole numbers by multiplying both sides of the equation by the least common denominator (LCD) of all the fractions, using the Multiplication Property.

For example,

[latex]\displaystyle{\frac{2}{3}x = \frac{5}{2} + 4}[/latex] Since the LCD of the denominators 3 and 2 is 6, multiply both sides of the equation by 6.

[latex]\displaystyle{6\left(\frac{2}{3}x\right) = 6\left(\frac{5}{2} + 4\right)}[/latex] This is the same as multiplying each term by the LCD of 6.

[latex]\displaystyle{6\left(\frac{2}{3}x\right) = 6\left(\frac{5}{2}\right) + 6(4)}[/latex] Simplifying, [latex]4x = 15 + 24[/latex] Now, the equation has only whole number coefficients [latex]4x = 39[/latex].

Equations with Decimal Coefficients

If an equation contains decimal coefficients, then the decimal coefficients can be changed to whole numbers by multiplying both sides of the equation by an appropriate power of 10, using the Multiplication Property.

[latex]\underline{1.25}x = \underline{0.2} + 4[/latex] Since there is at most 2 decimal places in any of the coefficients or constants, multiply both sides of the equation by [latex]10^2 = 100[/latex].

[latex]100(1.25x) = 100(0.2 + 4)[/latex] This is the same as multiplying each term by 100. [latex]100(1.25x) = 100(0.2) + 100(4)[/latex] Simplifying, [latex]125x = 20 + 400[/latex] Now, the equation has only whole number coefficients [latex]125x = 420[/latex].

Steps to Solve Algebraic Equations with One Variable

If the equation contains fraction and/or decimal coefficients, it is possible to work with them as they are – in that case, proceed to Step 2. Alternatively, as explained earlier, the equation may be rewritten in whole numbers to make calculations and rearrangements easier.

If present, expand and clear brackets in the equation by following the order of arithmetic operations (BEDMAS).

Use the addition and subtraction properties to collect and group all variable terms on the left side of the equation and all constants on the right side . Then, simplify both sides.

Note: If it is more convenient to gather all the variable terms on the right side and the constants on the left side, you may do so, and then use the symmetric property and switch the sides of the equation to bring the variables over to the left side and the constants to the right side.

Use the division and multiplication properties to ensure that the coefficient of the variable is +1.

After completing Step 4, there should be a single variable with a coefficient of +1 on the left side and a single constant term on the right side – that constant term is the solution to the equation.

Verify the answer by substituting the solution from Step 5 back into the original problem.

State the answer.

Example 7.3-a: Solving Equations Using the Addition and Subtraction Properties

Solve the following equations and verify the solutions:

  • [latex]x - 11 = 4[/latex]
  • [latex]8 + x = 20[/latex]
  • [latex]x - 11 = 4[/latex] Adding 11 to both sides, [latex]x - 11 + 11 = 4 + 11[/latex] [latex]x = 15[/latex] Verify by substituting [latex]x = 15[/latex]: LS [latex]= x - 11 = 15 - 11 = 4[/latex]     RS [latex]= 4[/latex] LS = RS Therefore, the solution is [latex]x = 15[/latex].
  • [latex]8 + x = 20[/latex] Subtracting 8 from both sides, [latex]8 - 8 + x = 20 - 8[/latex] [latex]x = 12[/latex] Verify by substituting [latex]x = 12[/latex]: LS [latex]= 8 + x = 8 + 12 = 20[/latex]   RS [latex]= 20[/latex] LS = RS Therefore, the solution is [latex]x = 12[/latex].

Example 7.3-b: Solving Equations Using the Multiplication and Division Properties

  • [latex]5x = 20[/latex]
  • [latex]\displaystyle{\frac{3}{8}x = 12}[/latex]
  • [latex]5x = 20[/latex] Dividing both sides by 5 , [latex]\displaystyle{\frac{5x}{5} = \frac{20}{5}}[/latex] [latex]x = 4[/latex] Verify by substituting [latex]x = 4[/latex]: LS [latex]= 5x = 5(4) = 20[/latex]   RS [latex]= 20[/latex] LS = RS Therefore, the solution is [latex]x = 4[/latex].
  • [latex]\displaystyle{\frac{3}{8}x = 12}[/latex] Multiplying both sides by [latex]\displaystyle{\frac{8}{3}}[/latex] (the reciprocal of [latex]\displaystyle{\frac{3}{8}}[/latex]), [latex]\displaystyle{\left(\frac{8}{3}\right) \cdot \frac{3}{8}x = \left(\frac{8}{3}\right) \cdot 12}[/latex] [latex]x = 8 \times 4[/latex] [latex]x = 32[/latex]   or  [latex]\displaystyle{\frac{3}{8}x = 12}[/latex] Multiplying both sides by 5 , [latex]\displaystyle{(8) \cdot \frac{3}{8}x = (8) \cdot 12}[/latex] [latex]3x = 96[/latex]Dividing both sides by 3 , [latex]\displaystyle{\frac{3x}{3} = \cdot \frac{96}{3}}[/latex][latex]x = 32[/latex]Verify by substituting [latex]x = 32[/latex]:LS [latex]\displaystyle{= \frac{3}{8}x = \frac{3}{8} \times 32 = 12}[/latex]RS [latex]= 12[/latex]LS = RSTherefore, the solution is [latex]x = 32[/latex].

Example 7.3-c: Solving Equations with Variables on Both Sides

  • [latex]3x - 8 = 12 - 2x[/latex]
  • [latex]15 + 6x - 4 = 3x + 31 - x[/latex]
  • [latex]3x - 8 = 12 - 2x[/latex] Adding 2x to both sides,[latex]3x + 2x - 8 = 12 - 2x + 2x[/latex] [latex]5x - 8 = 12[/latex] Adding 8 to both sides,[latex]5x - 8 + 8 = 12 + 8[/latex] [latex]5x = 20[/latex] Dividing both sides by 5 , [latex]\displaystyle{\frac{5x}{5} = \frac{20}{5}}[/latex] [latex]x = 4[/latex] Verify by substituting [latex]x = 4[/latex]: LS [latex]= 3x - 8 = 3(4) - 8 = 12 - 8 = 4[/latex]RS [latex]= 12 - 2x = 12 - 2(4) = 12 - 8 = 4[/latex]LS = RSTherefore, the solution is [latex]x = 4[/latex].
  • [latex]15 + 6x - 4 = 3x + 31 - x[/latex] Combining like terms (LS: [latex]15 - 4 = 11[/latex], and RS: [latex]3x - x = 2x[/latex]), [latex]11 + 6x = 2x + 31[/latex]   Subtracting [latex]2x[/latex] from both sides, [latex]11 + 6x - 2x = 2x - 2x + 31[/latex] [latex]11 + 4x = 31[/latex] Subtracting 11 from both sides, [latex]11 - 11 + 4x = 31 - 11[/latex] [latex]4x = 20[/latex] Dividing both sides by 4 , [latex]\displaystyle{\frac{4x}{4} = \frac{20}{4}}[/latex] [latex]x = 5[/latex] Verify by substituting [latex]x = 5[/latex] back into the original equation:LS [latex]= 15 + 6x - 4 = 15 + 6(5) - 4 = 15 + 30 - 4 = 41[/latex]RS [latex]= 3x + 31 - x = 3(5) + 31 - 5 = 15 + 31 - 5 = 41[/latex]LS = RSTherefore, the solution is [latex]x = 5[/latex].

Example 7.3-d: Solving Equations with Fractions

Solve the following equation and verify the solution:

[latex]\displaystyle{\frac{x}{3} - \frac{1}{12} = \frac{1}{6} + \frac{x}{4}}[/latex]

[latex]\displaystyle{\frac{x}{3} - \frac{1}{12} = \frac{1}{6} + \frac{x}{4}}[/latex] LCD of 3, 4, 6, and 12 is 12.

Multiplying each term by 12 ,

[latex]\displaystyle{12\left(\frac{x}{3}\right) - 12\left(\frac{1}{12}\right) = 12\left(\frac{1}{6}\right) + 12\left(\frac{x}{4}\right)}[/latex]

[latex]4x - 1 = 2 + 3x[/latex]

Subtracting [latex]3x[/latex] from both sides,

[latex]4x - 3x - 1 = 2 + 3x - 3x[/latex]

[latex]x - 1 = 2[/latex]

Adding 1 to both sides,

[latex]x - 1 + 1 = 2 + 1[/latex]

[latex]x = 3[/latex]

Verify by substituting [latex]x = 3[/latex] back into the original equation:

LS [latex]\displaystyle{= \frac{x}{3} - \frac{1}{12} = \frac{3}{3} - \frac{1}{12} = \frac{12}{12} - \frac{1}{12} = \frac{11}{12}}[/latex]

RS [latex]\displaystyle{= \frac{1}{6} + \frac{x}{4} = \frac{1}{6} + \frac{3}{4} = \frac{2}{12} + \frac{9}{12} = \frac{11}{12}}[/latex]

Therefore, the solution is [latex]x = 3[/latex].

Example 7.3-e: Solving Equations with Decimals

[latex]0.15x + 1.2 = 0.4x - 0.8[/latex]

[latex]0.15x + 1.2 = 0.4x - 0.8[/latex] Greatest number of decimal places is 2 (i.e., hundredths).

Multiplying all the terms by [latex]10^2 = 100[/latex],

[latex]100(0.15x) + 100(1.2) = 100(0.4x) - 100(0.8)[/latex]

[latex]15x + 120 = 40x - 80[/latex]

Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS,

[latex]40x - 80 = 15x + 120[/latex]

Subtracting [latex]15x[/latex] from both sides,

[latex]40x - 15x - 80 = 15x - 15x + 120[/latex]

[latex]25x - 80 = 120[/latex]

Adding 80 to both sides,

[latex]25x - 80 + 80 = 120 + 80[/latex]

[latex]25x = 200[/latex]

Dividing both sides by 25 ,

[latex]\displaystyle{\frac{25x}{25} = \frac{200}{25}}[/latex]

[latex]x = 8[/latex]

Verify by substituting [latex]x = 8[/latex] back into the original equation:

LS [latex]= 0.15x + 1.2 = 0.15(8) + 1.2 = 1.2 + 1.2 = 2.4[/latex]

RS [latex]= 0.4x - 0.8 = 0.4(8) - 0.8 = 3.2 - 0.8 = 2.4[/latex]

Therefore, the solution is [latex]x = 8[/latex].

Note: For the rest of the examples in this section, we will not show the verification by substitution step.

Example 7.3-f: Solving Equations Using All the Properties

Solve the following equations by using the properties of equality, and express the answer as a fraction in its lowest terms, or as a mixed number, wherever applicable:

  • [latex]8x + 7 - 3x = -6x - 15 + x[/latex]
  • [latex]2(3x - 7) = 28 - 3(x + 1)[/latex]
  • [latex]\displaystyle{\frac{1}{4}(x + \frac{2}{3}) = \frac{1}{2}(x - 3) + x}[/latex]
  • [latex]0.45(2x + 3) - 2.55 = 0.6(3x - 5)[/latex]
  • [latex]\displaystyle{\frac{x + 2}{3} = \frac{5 - 2x}{7}}[/latex]
  • [latex]8x + 7 - 3x = -6x - 15 + x[/latex] Grouping like terms on both sides, [latex]8x - 3x + 7 = -6x + x - 15[/latex] [latex]5x + 7 = -5x - 15[/latex] Adding 5 to both sides, [latex]5x + 5x + 7 = -5x + 5x - 15[/latex] [latex]10x + 7 = -15[/latex] Subtracting 7 from both sides, [latex]10x + 7 - 7 = -15 - 7[/latex] [latex]10x = -22[/latex] Dividing both sides by 10 , [latex]\displaystyle{\frac{10x}{10} = -\frac{22}{10}}[/latex] [latex]\displaystyle{x = -\frac{11}{5} = -2\frac{1}{5}}[/latex]
  • [latex]2(3x - 7) = 28 - 3(x + 1)[/latex] Expanding both sides,[latex]6x - 14 = 28 - 3x - 3[/latex]   Grouping like terms,[latex]6x - 14 = 28 - 3 - 3x[/latex] [latex]6x - 14 = 25 - 3x[/latex] Adding [latex]3x[/latex] to both sides,[latex]6x + 3x - 14 = 25 - 3x + 3x[/latex] [latex]9x - 14 = 25[/latex] Adding 14 to both sides, [latex]9x - 14 + 14 = 25 + 14[/latex] [latex]9x = 39[/latex] Dividing both sides by 9 ,[latex]\displaystyle{\frac{9x}{9} = \frac{39}{9}}[/latex] [latex]\displaystyle{x = \frac{13}{3} = 4\frac{1}{3}}[/latex]
  • [latex]\displaystyle{\frac{1}{4}(x + \frac{2}{3}) = \frac{1}{2}(x - 3) + x}[/latex] Expanding both sides, [latex]\displaystyle{\frac{1}{4}x + \frac{1}{6} = \frac{1}{2}x - \frac{3}{2} + x}[/latex] Multiplying each term by the LCD 12 , [latex]\displaystyle{12\left(\frac{1}{4}x\right) + 12\left(\frac{1}{6}\right) = 12\left(\frac{1}{2}x\right) - 12\left(\frac{3}{2}\right) + 12(x)}[/latex] [latex]3x + 2 = 6x - 18 + 12x[/latex] Grouping like terms,[latex]3x + 2 = 6x + 12x - 18[/latex] [latex]3x + 2 = 18x - 18[/latex] Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS, [latex]18x 18 = 3x + 2[/latex]   Subtracting [latex]3x[/latex] from both sides, [latex]18x - 3x - 18 = 3x - 3x + 2[/latex] [latex]15x - 18 = 2[/latex] Adding 18 to both sides, [latex]15x - 18 + 18 = 2 + 18[/latex] [latex]15x = 20[/latex] Dividing both sides by 15 , [latex]\displaystyle{\frac{15x}{15} = \frac{20}{15}}[/latex][latex]\displaystyle{x = \frac{4}{3} = 1\frac{1}{3}}[/latex]
  • [latex]0.45(2x + 3) - 2.55 = 0.6(3x - 5)[/latex] Expanding both sides, [latex]0.90x + 1.35 - 2.55 = 1.8x - 3.0[/latex] Greatest number of decimal places is 2 (i.e., hundredths). Multiplying all the terms by [latex]10^2 = 100[/latex],   [latex]100(0.90x) + 100(1.35) - 100(2.55) = 100(1.8x) - 100(3.0)[/latex] [latex]90x + 135 - 255 = 180x - 300[/latex] Grouping like terms,[latex]90x 120 = 180x 300[/latex] Interchanging the LS and RS using the Symmetric Property to have the larger x term on the LS,   [latex]180x - 300 = 90x - 120[/latex] Subtracting [latex]90x[/latex] from both sides,  [latex]180x - 90x - 300 = 90x - 90x - 120[/latex] [latex]90x - 300 = -120[/latex] Adding 300 to both sides,  [latex]90x - 300 + 300 = -120 + 300[/latex] [latex]90x = 180[/latex] Dividing both sides by 90 ,  [latex]\displaystyle{\frac{90x}{90} = \frac{180}{90}}[/latex] [latex]x = 2[/latex]
  • [latex]\displaystyle{\frac{x + 2}{3} = \frac{5 - 2x}{7}}[/latex] Cross-multiplying,  [latex]7(x + 2) = 3(5 - 2x)[/latex] Expanding both sides,[latex]7x + 14 = 15 - 6x[/latex] Adding [latex]6x[/latex] to both sides,  [latex]7x + 6x + 14 = 15 - 6x + 6x[/latex] [latex]13x + 14 = 15[/latex] Subtracting 14 from both sides, [latex]13x + 14 - 14 = 15 - 14[/latex] [latex]13x = 1[/latex] Dividing both sides by 13 ,  [latex]\displaystyle{\frac{13x}{13} = \frac{1}{13}}[/latex] [latex]\displaystyle{x = \frac{1}{13}}[/latex]

Steps to Solve Word Problems

Read the entire problem and ensure you understand the situation.

Identify the given information and the question to be answered.

Look for keywords. Some words indicate certain mathematical operations (see Table 7.1).

Choose a variable to represent the unknown(s) and state what that variable represents, including the unit of measure.

Note: For now, if there is more than one unknown, try to identify all the unknowns in terms of one variable, as all the questions in this chapter can be solved with only one variable.

Where necessary, draw a simple sketch to identify the information. This helps with envisioning the question more clearly.

Create an equation (or set of equations) to describe the relationship between the variables and the constants in the question.

Group like terms, isolate the variable and solve for the unknown(s).

State the solution to the given problem.

Example 7.3-g: Solving a Word Problem Using Algebraic Equations

If Harry will be 65 years old in 5 years, how old is he today?

Let Harry’s age today be x years.

Therefore, in 5 years, Harry’s age will be:

[latex]x + 5 = 65[/latex]   Solving for [latex]x[/latex],

[latex]x = 65 - 5 = 60[/latex]

Therefore, Harry is 60 years old today.

Example 7.3-h: Solving a Geometry Problem Using Algebraic Equations

The perimeter of a rectangular garden is 50 metres. The length is 5 metres more than the width. Find the dimensions of the garden.

Hint: Perimeter = 2(length) + 2(width)

Let the width be [latex]w[/latex] metres.

Therefore, the length is (w + 5) metres.

Perimeter = 2(length) + 2(width)

[latex]50 = 2(w + 5) + 2w[/latex]

[latex]50 = 2w + 10 + 2w[/latex]

[latex]2w + 10 + 2w = 50[/latex]

[latex]4w + 10 = 50[/latex]

[latex]4w = 50 - 10[/latex]

[latex]4w = 40[/latex]

[latex]\displaystyle{w = \frac{40}{4}}[/latex]

[latex]w = 10[/latex]

Therefore, the width of the garden is 10 metres and the length is (10 + 5) = 15 metres.

Example 7.3-i: Solving a Finance Problem Using Algebraic Equations

A TV costs $190 more than a Blu-ray player. The total cost of the TV and the Blu-ray player is $688. Calculate the cost of the TV and the cost of the Blu-ray player.

Let the cost of the Blu-ray player be [latex]\$x[/latex].

Therefore, the cost of the TV is [latex]\$(x + 190.00)[/latex].

The total cost is $688.00.

[latex]x + (x + 190.00) = 688.00[/latex]

[latex]x + x + 190.00 = 688.00[/latex]

[latex]2x + 190.00 = 688.00[/latex]

[latex]2x = 688.00 - 190.00[/latex]

[latex]2x = 498.00[/latex]

[latex]\displaystyle{x = \frac{498.00}{2}}[/latex]

[latex]x = \$249.00[/latex]

Therefore, the cost of the Blu-ray player is $249.00 and the cost of the TV is (249.00 + 190.00) = $439.00.

Example 7.3-j: Solving a Mixture Problem Using Algebraic Equations

How many litres of water need to be added to 30 litres of a 15% saline solution to make a saline solution that is 10% saline?

From the last column, we get the equation for the saline mix. The number of litres of saline in the 15% solution must be the same as the number of litres in the final 10% solution, as only water is being added, which does not contribute any additional saline to the solution. Therefore,

[latex]4.5 = 0.10 \times (30 + x)[/latex]

[latex]4.5 = 3 + 0.10x[/latex]

[latex]1.5 = 0.10x[/latex]

[latex]x = 15[/latex]

Therefore, 15 litres of water need to be added to the 15% saline solution to make the solution 10% saline.

7.3 Exercises

Answers to the odd-numbered problems are available at the end of the textbook .

For problems 1 to 8, simplify and evaluate the expressions.

  • The sum of a number and six is ten.
  • A number decreased by fifteen is five.
  • Six times a number is seventy-two.
  • The product of a number and four is twenty-eight.
  • A number divided by five is four.
  • A number divided by three is three.
  • Two-thirds of a number is twelve.
  • Two-fifths of a number is six.

For problems 9 to 30, solve the algebraic equations using the properties of equality, and express the answer as a fraction in its lowest terms or as a mixed number, wherever applicable.

  • [latex]x - 20 = 10[/latex]
  • [latex]x - 25 = 17[/latex]
  • [latex]22 = 40 - x[/latex]
  • [latex]54 = 23 - x[/latex]
  • [latex]21 + x = 4[/latex]
  • [latex]50 + x = 45[/latex]
  • [latex]16 + x = 22[/latex]
  • [latex]12 + x = 38[/latex]
  • [latex]11x + 4 = 17[/latex]
  • [latex]7x - 16 = 22[/latex]
  • [latex]\displaystyle{x - \frac{4}{5} = \frac{3}{5}}[/latex]
  • [latex]\displaystyle{x - \frac{1}{6} = 1}[/latex]
  • [latex]\displaystyle{\frac{10}{15} = x - \frac{4}{3}}[/latex]
  • [latex]\displaystyle{\frac{x}{7} + 15 = 24}[/latex]
  • [latex]\displaystyle{x + \frac{2}{5} = \frac{1}{4}}[/latex]
  • [latex]\displaystyle{2x - \frac{2}{3} = \frac{5}{6}}[/latex]
  • [latex]4x = 24[/latex]
  • [latex]\displaystyle{\frac{2x}{3} + 1 = \frac{5x}{8} + 2}[/latex]
  • [latex]\displaystyle{\frac{x}{2} - \frac{1}{6} = \frac{1}{3} + \frac{3x}{5}}[/latex]
  • [latex]\displaystyle{\frac{7x}{8} - 4 = \frac{x}{4} + 6}[/latex]
  • [latex]\displaystyle{\frac{8x}{3} - 5 = \frac{x}{3} + 2}[/latex]

For problems 31 to 54, solve the algebraic equations using the properties of equality, and round the answer to 2 decimal places, wherever applicable.

  • [latex]10y - 0.09y = 17[/latex]
  • [latex]x + 0.13x = 70[/latex]
  • [latex]0.3x - 3.2 = 0.4 - 0.6x[/latex]
  • [latex]4 + 0.2x = 0.7x - 0.5[/latex]
  • [latex]0.4x - 1.38 = 0.3x - 1.2[/latex]
  • [latex]1.2 - 0.7x = 2.7 - 0.5x[/latex]
  • [latex]0.43x + 0.25 = 0.29x - 0.03[/latex]
  • [latex]0.6x - 1.2 = 0.9 - 1.5x[/latex]
  • [latex]x - 2 - 4x = -3x - 8 + 5x[/latex]
  • [latex]4(2x - 5) = 32 - 4(x - 2)[/latex]
  • [latex](4 + 6)(2 + 4x) = 45 - 2.5(x + 3)[/latex]
  • [latex](5 + 0.5x)(1 + 3) = -1.2(2x + 4) + 25[/latex]
  • [latex]15 + 5(x - 10) = 3(x - 1)[/latex]
  • [latex]2(x - 3) + 3(x - 5) = 4[/latex]
  • [latex]4(y + 7) - 2(y - 4) = 3(y - 2)[/latex]
  • [latex]8(2y + 4) - 6(3y + 7) = 3y[/latex]
  • [latex]\displaystyle{\frac{x - 7}{2} + \frac{x + 2}{3} = 41}[/latex]
  • [latex]\displaystyle{\frac{7}{12}(2x + 1) + \frac{3}{4}(x + 1) = 3}[/latex]
  • [latex]\displaystyle{\frac{5}{y + 4} = \frac{3}{y - 2}}[/latex]
  • [latex]\displaystyle{\frac{3}{x + 1} = \frac{2}{x - 3}}[/latex]
  • [latex]\displaystyle{\frac{7}{5x - 3} = \frac{5}{4x}}[/latex]
  • [latex]\displaystyle{\frac{5}{y + 2} = \frac{3}{y}}[/latex]

For problems 55 to 76, solve the word problems using algebraic equations.

  • If three times a number plus twenty is seven times that number, what is the number?
  • Fifteen less than three times a number is twice that number. What is the number?
  • A 25-metre-long wire is cut into two pieces. One piece is 7 metres longer than the other. Find the length of each piece.
  • A 9-metre-long pipe is cut into two pieces. One piece is twice the length of the other piece. Find the length of each piece.
  • $500 is shared between Andy and Becky. Andy’s share is $150 less than Becky’s share. Calculate the amount of each of their shares.
  • $200 is shared between Bill and Ann. Ann’s share is $50 more than Bill’s share. Calculate the size of each of their shares.
  • Movie tickets that were sold to each child were $3 cheaper than those sold to each adult. If a family of two adults and two children paid $34 to watch a movie at the cinema, what was the price of each adult ticket and each child ticket?
  • Giri had twice the number of quarters (25 cents) in his bag than dimes (10 cents). If he had a total of 54 coins, how many of them were quarters? What was the total dollar value of these coins?
  • A square garden, with sides of length x metres, is widened by 4 metres and lengthened by 3 metres. Write the equation for the area (A) of the expanded garden. If each side was originally 10 metres in length, find the new area. (Hint: Area of a Rectangle = Length × Width)
  • A square garden, with sides of length x metres, has had its width reduced by 4 metres and its length reduced by 2 metres. Write the equation for the Area (A) of the smaller garden. If each side was originally 20 metres in length, find the new area.
  • Aran bought a shirt and a pair of pants for $34.75. The pair of pants cost $9.75 more than the shirt. Calculate the cost of the shirt.
  • Mythili bought a schoolbag and a toy for $30.45. The school bag cost $5.45 more than the toy. Calculate the cost of the school bag.
  • Sam is paid $720 a week. He worked 9 hours of overtime last week and he received $954. Calculate his overtime rate per hour.
  • Lisa is paid $840 a week. Her overtime rate is $28 per hour. Last week she received $1,036. How many hours of overtime did she work last week?
  • The sum of the three angles of any triangle is 180°. If [latex]3x[/latex], [latex]7x[/latex], and [latex]8x[/latex] are the measures of the three angles of a triangle, calculate the measure of each angle of the triangle.
  • The sum of the three angles of any triangle is 180°. If [latex]3x[/latex], [latex]4x[/latex], and [latex]5x[/latex] are the measures of the three angles of a triangle, calculate the measure of each angle of the triangle.
  • The perimeter of a triangle is the sum of the lengths of the three sides of the triangle. The perimeter of a triangle with sides [latex]x[/latex] cm, [latex](x + 10)[/latex] cm, and [latex]2x[/latex] cm is 70 cm. Calculate the length of each side of the triangle.
  • The perimeter of a triangle is the sum of the lengths of the three sides of the triangle. The perimeter of a triangle with sides [latex]x + 10[/latex], [latex]2x + 10[/latex], and [latex]3x[/latex] is 110 cm. Calculate the length of each side of the triangle.
  • After completing a weight-loss program, a patient weighs 160 lb. His dietician observes that the patient has lost 15% of his original weight. What was the patient’s starting weight?
  • A beaker in a chemistry lab contains 3 litres of water. While conducting an experiment, the chemistry professor removes three-fifths of the water from the beaker. He then adds three-fifths of the remaining volume to the beaker. How much water is left in the beaker at the end of the experiment?
  • A researcher wants to make 4 L of a 7% acid solution. She has a beaker of 15% acid solution in stock. How much of the 15% solution does she need to use and how much water must she add in order to prepare her desired solution?
  • A chemist wants to make a 10% acid solution. She has 5 L of 25% acid solution. How many litres of water should she add to the 25% solution in order to prepare her desired solution?

Unless otherwise indicated, this chapter is an adaptation of the eTextbook  Foundations of Mathe matics  (3 rd  ed.) by Thambyrajah Kugathasan, published by  Vretta-Lyryx Inc ., with permission. Adaptations include supplementing existing material and reordering chapters.

Fundamentals of Business Math Copyright © 2023 by Lisa Koster and Tracey Chase is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

Steps For Solving Real World Problems

  • Highlight the important information in the problem that will help write two equations.
  • Define your variables
  • Write two equations
  • Use one of the methods for solving systems of equations to solve.
  • Check your answers by substituting your ordered pair into the original equations.
  • Answer the questions in the real world problems. Always write your answer in complete sentences!

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1.  Let's start by identifying the important information:

  • hot dogs cost $1.50
  • Sodas cost $0.50
  • Made a total of $78.50
  • Sold 87 hot dogs and sodas combined

2.  Define your variables.

  • Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50    (Equation related to cost)

 x + y = 87   (Equation related to the number sold)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

Solving a systems using substitution

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6.  Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7.  Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

  • 3 soft tacos + 3 burritos cost $11.25
  • 4 soft tacos + 2 burritos cost $10.00

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25  (Equation representing your lunch)

4x + 2y = 10   (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

Solving Systems Using Combinations

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

solving equations word problem examples

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

  • System of Equations
  • Systems Word Problems

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Calcworkshop

Solving Linear Equation Word Problems 9 Terrific Examples!

// Last Updated: January 20, 2020 - Watch Video //

What’s the first thing that comes to mind when you hear the phrase Word problems?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching linear word problems

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

For some, it’s a chance to solve a real-world example, so there’s a level of excitement and sense of wonder. For others, it’s groaning, and frustration on where to even begin.

Well, in this lesson we’re going to make Solving Linear Equation Word Problems manageable with easy to follow tricks and steps.

We already know how to solve all different types of equations. Yay!

And we also know how to translate algebraic expressions and equations. Double Yay!

Now it’s time to bring both of these together.

Finding the length of the missing side of a triangle

Solving equations and word problems Example

But, what about the tricks and steps?

Yes, there are some easy to follow steps that we are going to use to solve linear word problems.

  • Read the problem carefully and determine what is being asked.
  • Create a sidebar! Using different colors, symbols and diagrams and write an equation the relates all the information given.
  • Solve your equation and check your answer(s).

Now, these steps might not seem all that remarkable, but once you see them in action I guarantee that writing equations from word problems and solving them will become like second nature!

Again, the secret to success is your Sidebar. This is where you will write down all the information you’ve gleaned from the problem, and formulate a solution by writing an equation to model the situation, as Khan Academy accurately states.

Together we will walk through 9 examples in detail ranging from finding consecutive integers to finding hourly wages, profit and cost, distances for rectangles and triangles, and people’s ages.

Linear Word Problems (How-To) – Video

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Polynomial Equation Word Problems (Mixed Operations)

These lessons help Algebra students learn how to write and solve polynomial equations for algebra word problems.

Related Pages Solving Challenging Word Problems Math Word Problems More Algebra Lessons

How To Solve Polynomial Equation Word Problem?

Example: A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

Polynomial Equation Word Problem

Example: A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air as follows: h = -16t 2 + 8t + 8 a) How long will it take the gymnast to reach the ground? b) When will the gymnast be 8 feet above the ground?

How To Solve Word Problems With Polynomial Equations?

  • The sum of a number and its square is 72. Find the number.
  • The area of a triangle is 44m 2 . Find the lengths of the legs if one of the legs is 3m longer than the other leg.
  • The top of a 15-foot ladder is 3 feet farther up a wall than the foo is from the bottom of the wall. How far is the ladder from the bottom of the wall?
  • A projectile is launched upward from ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground?

How To Write Polynomials For Word Problems?

Learn to write a polynomial for Word problems involving perimeter and area of rectangles and circles.

Learn How To Write And Solve Polynomial Equations

Learn to write and solve polynomial equations for special integers, consecutive integers.

Example 1: Find a number that is 56 less than its square. Let n be the number. Example 2: Find two consecutive odd integers whose sum is 130.

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Chapter 6: Polynomials

6.8 Mixture and Solution Word Problems

Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below:

The first column in the table (Name) is used to identify the fluids or objects being mixed in the problem. The second column (Amount) identifies the amounts of each of the fluids or objects. The third column (Value) is used for the value of each object or the percentage of concentration of each fluid. The last column (Equation) contains the product of the Amount times the Value or Concentration.

Example 6.8.1

Jasnah has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Find the equation.

  • The solution names are 50% (S 50 ), 60% (S 60 ), and 80% (S 80 ).
  • The amounts are S 50 = 70 mL, S 80 , and S 60 = 70 mL + S 80 .
  • The concentrations are S 50 = 0.50, S 60 = 0.60, and S 80 = 0.80.

The equation derived from this data is 0.50 (70 mL) + 0.80 (S 80 ) = 0.60 (70 mL + S 80 ).

Example 6.8.2

Sally and Terry blended a coffee mix that sells for [latex]\$2.50[/latex] by mixing two types of coffee. If they used 40 mL of a coffee that costs [latex]\$3.00,[/latex] how much of another coffee costing [latex]\$1.50[/latex] did they mix with the first?

The equation derived from this data is:

[latex]\begin{array}{ccccccc} 1.50(C_{1.50})&+&3.00(40)&=&2.50(40&+&C_{1.50}) \\ 1.50(C_{1.50})&+&120&=&100&+&2.50(C_{1.50}) \\ -2.50(C_{1.50})&-&120&=&-120&-&2.50(C_{1.50}) \\ \hline &&-1.00(C_{1.50})&=&-20&& \\ \\ &&(C_{1.50})&=&\dfrac{-20}{-1}&& \\ \\ &&C_{1.50}&=&20&& \end{array}[/latex]

This means 20 mL of the coffee selling for [latex]\$1.50[/latex] is needed for the mix.

Example 6.8.3

Nick and Chloe have two grades of milk from their small dairy herd: one that is 24% butterfat and another that is 18% butterfat. How much of each should they use to end up with 42 litres of 20% butterfat?

The equation derived from this data is:

[latex]\begin{array}{rrrrrrr} 0.24(B_{24})&+&0.18(42&- &B_{24})&=&0.20(42) \\ 0.24(B_{24})&+&7.56&-&0.18(B_{24})&=&8.4 \\ &-&7.56&&&&-7.56 \\ \hline &&&&0.06(B_{24})&=&0.84 \\ \\ &&&&B_{24}&=&\dfrac{0.84}{0.06} \\ \\ &&&&B_{24}&=&14 \end{array}[/latex]

This means 14 litres of the 24% buttermilk, and 28 litres of the 18% buttermilk is needed.

Example 6.8.4

In Natasha’s candy shop, chocolate, which sells for [latex]\$4[/latex] a kilogram, is mixed with nuts, which are sold for [latex]\$2.50[/latex] a kilogram. Chocolate and nuts are combined to form a chocolate-nut candy, which sells for [latex]\$3.50[/latex] a kilogram. How much of each are used to make 30 kilograms of the mixture?

[latex]\begin{array}{rrrrrrl} 4.00(C)&+&2.50(30&-&C)&=&3.50(30) \\ 4.00(C)&+&75&-&2.50(C)&=&105 \\ &-&75&&&&-75 \\ \hline &&&&1.50(C)&=&30 \\ \\ &&&&C&=&\dfrac{30}{1.50} \\ \\ &&&&C&=&20 \end{array}[/latex]

Therefore, 20 kg of chocolate is needed for the mixture.

With mixture problems, there is often mixing with a pure solution or using water, which contains none of the chemical of interest. For pure solutions, the concentration is 100%. For water, the concentration is 0%. This is shown in the following example.

Example 6.8.5

Joey is making a a 65% antifreeze solution using pure antifreeze mixed with water. How much of each should be used to make 70 litres?

[latex]\begin{array}{rrrrl} 1.00(A)&+&0.00(70-A)&=&0.65(0.70) \\ &&1.00A&=&45.5 \\ &&A&=&45.5 \\ \end{array}[/latex]

This means the amount of water added is 70 L − 45.5 L = 24.5 L.

For questions 1 to 9, write the equations that define the relationship.

  • A tank contains 8000 litres of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid?
  • How much pure antifreeze should be added to 5 litres of a 30% mixture of antifreeze to make a solution that is 50% antifreeze?
  • You have 12 kilograms of 10% saline solution and another solution of 3% strength. How many kilograms of the second should be added to the first in order to get a 5% solution?
  • How much pure alcohol must be added to 24 litres of a 14% solution of alcohol in order to produce a 20% solution?
  • How many litres of a blue dye that costs [latex]\$1.60[/latex] per litre must be mixed with 18 litres of magenta dye that costs [latex]\$2.50[/latex] per litre to make a mixture that costs [latex]\$1.90[/latex] per litre?
  • How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?
  • A 100-kg bag of animal feed is 40% oats. How many kilograms of pure oats must be added to this feed to produce a blend of 50% oats?
  • A 20-gram alloy of platinum that costs [latex]\$220[/latex] per gram is mixed with an alloy that costs [latex]\$400[/latex] per gram. How many grams of the [latex]\$400[/latex] alloy should be used to make an alloy that costs [latex]\$300[/latex] per gram?
  • How many kilograms of tea that cost [latex]\$4.20[/latex] per kilogram must be mixed with 12 kilograms of tea that cost [latex]\$2.25[/latex] per kilogram to make a mixture that costs [latex]\$3.40[/latex] per kilogram?

Solve questions 10 to 21.

  • How many litres of a solvent that costs [latex]\$80[/latex] per litre must be mixed with 6 litres of a solvent that costs [latex]\$25[/latex] per litre to make a solvent that costs [latex]\$36[/latex] per litre?
  • How many kilograms of hard candy that cost [latex]\$7.50[/latex] per kg must be mixed with 24 kg of jelly beans that cost [latex]\$3.25[/latex] per kg to make a mixture that sells for [latex]\$4.50[/latex] per kg?
  • How many kilograms of soil supplement that costs [latex]\$7.00[/latex] per kg must be mixed with 20 kg of aluminum nitrate that costs [latex]\$3.50[/latex] per kg to make a fertilizer that costs [latex]\$4.50[/latex] per kg?
  • A candy mix sells for [latex]\$2.20[/latex] per kg. It contains chocolates worth [latex]\$1.80[/latex] per kg and other candy worth [latex]\$3.00[/latex] per kg. How much of each are in 15 kg of the mixture?
  • A certain grade of milk contains 10% butterfat and a certain grade of cream 60% butterfat. How many litres of each must be taken so as to obtain a mixture of 100 litres that will be 45% butterfat?
  • Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100 cc of a solution that is 68% acid?
  • A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many litres of each must be used to make 600 litres of paint that is 19% green dye?
  • How many kilograms of coffee that is 40% java beans must be mixed with coffee that is 30% java beans to make an 80-kg coffee blend that is 32% java beans?
  • A caterer needs to make a slightly alcoholic fruit punch that has a strength of 6% alcohol. How many litres of fruit juice must be added to 3.75 litres of 40% alcohol?
  • A mechanic needs to dilute a 70% antifreeze solution to make 20 litres of 18% strength. How many litres of water must be added?
  • How many millilitres of water must be added to 50 millilitres of 100% acid to make a 40% solution?
  • How many litres of water need to be evaporated from 50 litres of a 12% salt solution to produce a 15% salt solution?

Answer Key 6.8

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Sat / act prep online guides and tips, the complete guide to sat math word problems.

feature_words-1

About 25% of your total SAT Math section will be word problems, meaning you will have to create your own visuals and equations to solve for your answers. Though the actual math topics can vary, SAT word problems share a few commonalities, and we’re here to walk you through how to best solve them.

This post will be your complete guide to SAT Math word problems. We'll cover how to translate word problems into equations and diagrams, the different types of math word problems you’ll see on the test, and how to go about solving your word problems on test day.

Feature Image: Antonio Litterio /Wikimedia

What Are SAT Math Word Problems?

A word problem is any math problem based mostly or entirely on a written description. You will not be provided with an equation, diagram, or graph on a word problem and must instead use your reading skills to translate the words of the question into a workable math problem. Once you do this, you can then solve it.

You will be given word problems on the SAT Math section for a variety of reasons. For one, word problems test your reading comprehension and your ability to visualize information.

Secondly, these types of questions allow test makers to ask questions that'd be impossible to ask with just a diagram or an equation. For instance, if a math question asks you to fit as many small objects into a larger one as is possible, it'd be difficult to demonstrate and ask this with only a diagram.

Translating Math Word Problems Into Equations or Drawings

In order to translate your SAT word problems into actionable math equations you can solve, you’ll need to understand and know how to utilize some key math terms. Whenever you see these words, you can translate them into the proper mathematical action.

For instance, the word "sum" means the value when two or more items are added together. So if you need to find the sum of a and b , you’ll need to set up your equation like this: a+b.

Also, note that many mathematical actions have more than one term attached, which can be used interchangeably.

Here is a chart with all the key terms and symbols you should know for SAT Math word problems:

Now, let's look at these math terms in action using a few official examples:

body_sat_math_sample_question_1

We can solve this problem by translating the information we're given into algebra. We know the individual price of each salad and drink, and the total revenue made from selling 209 salads and drinks combined. So let's write this out in algebraic form.

We'll say that the number of salads sold = S , and the number of drinks sold = D . The problem tells us that 209 salads and drinks have been sold, which we can think of as this:

S + D = 209

Finally, we've been told that a certain number of S and D have been sold and have brought in a total revenue of 836 dollars and 50 cents. We don't know the exact numbers of S and D , but we do know how much each unit costs. Therefore, we can write this equation:

6.50 S + 2 D = 836.5

We now have two equations with the same variables ( S and D ). Since we want to know how many salads were sold, we'll need to solve for D so that we can use this information to solve for S . The first equation tells us what S and D equal when added together, but we can rearrange this to tell us what just D equals in terms of S :

Now, just subtract S from both sides to get what D equals:

D = 209 − S

Finally, plug this expression in for D into our other equation, and then solve for S :

6.50 S + 2(209 − S ) = 836.5

6.50 S + 418 − 2 S = 836.5

6.50 S − 2 S = 418.5

4.5 S = 418.5

The correct answer choice is (B) 93.

body_sat_math_sample_question_2

This word problem asks us to solve for one possible solution (it asks for "a possible amount"), so we know right away that there will be multiple correct answers.

Wyatt can husk at least 12 dozen ears of corn and at most 18 dozen ears of corn per hour. If he husks 72 dozen at a rate of 12 dozen an hour, this is equal to 72 / 12 = 6 hours. You could therefore write 6 as your final answer.

If Wyatt husks 72 dozen at a rate of 18 dozen an hour (the highest rate possible he can do), this comes out to 72 / 18 = 4 hours. You could write 4 as your final answer.

Since the minimum time it takes Wyatt is 4 hours and the maximum time is 6 hours, any number from 4 to 6 would be correct.

body_Latin

Though the hardest SAT word problems might look like Latin to you right now, practice and study will soon have you translating them into workable questions.

Typical SAT Word Problems

Word problems on the SAT can be grouped into three major categories:

  • Word problems for which you must simply set up an equation
  • Word problems for which you must solve for a specific value
  • Word problems for which you must define the meaning of a value or variable

Below, we look at each world problem type and give you examples.

Word Problem Type 1: Setting Up an Equation

This is a fairly uncommon type of SAT word problem, but you’ll generally see it at least once on the Math section. You'll also most likely see it first on the section.

For these problems, you must use the information you’re given and then set up the equation. No need to solve for the missing variable—this is as far as you need to go.

Almost always, you’ll see this type of question in the first four questions on the SAT Math section, meaning that the College Board consider these questions easy. This is due to the fact that you only have to provide the setup and not the execution.

body_sat_math_sample_question_3

To solve this problem, we'll need to know both Armand's and Tyrone's situations, so let's look at them separately:

Armand: Armand sent m text messages each hour for 5 hours, so we can write this as 5m —the number of texts he sent per hour multiplied by the total number of hours he texted.

Tyrone: Tyrone sent p text messages each hour for 4 hours, so we can write this as 4 p —the number of texts he sent per hour multiplied by the total number of hours he texted.

We now know that Armand's situation can be written algebraically as 5m , and Tyrone's can be written as 4 p . Since we're being asked for the expression that represents the total number of texts sent by Armand and Tyrone, we must add together the two expressions:

The correct answer is choice (C) 5m + 4 p

Word Problem Type 2: Solving for a Missing Value

The vast majority of SAT Math word problem questions will fall into this category. For these questions, you must both set up your equation and solve for a specific piece of information.

Most (though not all) word problem questions of this type will be scenarios or stories covering all sorts of SAT Math topics , such as averages , single-variable equations , and ratios . You almost always must have a solid understanding of the math topic in question in order to solve the word problem on the topic.

body_sat_math_sample_question_4

Let's try to think about this problem in terms of x . If Type A trees produced 20% more pears than Type B did, we can write this as an expression:

x + 0.2 x = 1.2 x = # of pears produced by Type A

In this equation, x is the number of pears produced by Type B trees. If we add 20% of x (0.2 x ) to x , we get the number of pears produced by Type A trees.

The problem tells us that Type A trees produced a total of 144 pears. Since we know that 1.2 x is equal to the number of pears produced by Type A, we can write the following equation:

1.2 x = 144

Now, all we have to do is divide both sides by 1.2 to find the number of pears produced by Type B trees:

x = 144 / 1.2

The correct answer choice is (B) 120.

You might also get a geometry problem as a word problem, which might or might not be set up with a scenario, too. Geometry questions will be presented as word problems typically because the test makers felt the problem would be too easy to solve had you been given a diagram, or because the problem would be impossible to show with a diagram. (Note that geometry makes up a very small percentage of SAT Math . )

body_SAT_word_problem_5

This is a case of a problem that is difficult to show visually, since x is not a set degree value but rather a value greater than 55; thus, it must be presented as a word problem.

Since we know that x must be an integer degree value greater than 55, let us assign it a value. In this case, let us call x 56°. (Why 56? There are other values x could be, but 56 is guaranteed to work since it's the smallest integer larger than 55. Basically, it's a safe bet!)

Now, because x = 56, the next angle in the triangle—2 x —must measure the following:

Let's make a rough (not to scale) sketch of what we know so far:

body_triangle_ex_1

Now, we know that there are 180° in a triangle , so we can find the value of y by saying this:

y = 180 − 112 − 56

One possible value for y is 12. (Other possible values are 3, 6, and 9. )

Word Problem Type 3: Explaining the Meaning of a Variable or Value

This type of problem will show up at least once. It asks you to define part of an equation provided by the word problem—generally the meaning of a specific variable or number.

body_sat_math_sample_question_6

This question might sound tricky at first, but it's actually quite simple.

We know tha t P is the number of phones Kathy has left to fix, and d is the number of days she has worked in a week. If she's worked 0 days (i.e., if we plug 0 into the equation), here's what we get:

P = 108 − 23(0)

This means that, without working any days of the week, Kathy has 108 phones to repair. The correct answer choice, therefore, is (B) Kathy starts each week with 108 phones to fix.

body_juggle

To help juggle all the various SAT word problems, let's look at the math strategies and tips at our disposal.

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SAT Math Strategies for Word Problems

Though you’ll see word problems on the SAT Math section on a variety of math topics, there are still a few techniques you can apply to solve word problems as a whole.

#1: Draw It Out

Whether your problem is a geometry problem or an algebra problem, sometimes making a quick sketch of the scene can help you understand what exactly you're working with. For instance, let's look at how a picture can help you solve a word problem about a circle (specifically, a pizza):

body_sat_math_sample_question_7_2

If you often have trouble visualizing problems such as these, draw it out. We know that we're dealing with a circle since our focus is a pizza. We also know that the pizza weighs 3 pounds.

Because we'll need to solve the weight of each slice in ounces, let's first convert the total weight of our pizza from pounds into ounces. We're given the conversion (1 pound = 16 ounces), so all we have to do is multiply our 3-pound pizza by 16 to get our answer:

3 * 16 = 48 ounces (for whole pizza)

Now, let's draw a picture. First, the pizza is divided in half (not drawn to scale):

body_sat_math_sample_question_7_diagram_1

We now have two equal-sized pieces. Let's continue drawing. The problem then says that we divide each half into three equal pieces (again, not drawn to scale):

body_sat_math_sample_question_7_diagram_2

This gives us a total of six equal-sized pieces. Since we know the total weight of the pizza is 48 ounces, all we have to do is divide by 6 (the number of pieces) to get the weight (in ounces) per piece of pizza:

48 / 6 = 8 ounces per piece

The correct answer choice is (C) 8.

As for geometry problems, remember that you might get a geometry word problem written as a word problem. In this case, make your own drawing of the scene. Even a rough sketch can help you visualize the math problem and keep all your information in order.

#2: Memorize Key Terms

If you’re not used to translating English words and descriptions into mathematical equations, then SAT word problems might be difficult to wrap your head around at first. Look at the chart we gave you above so you can learn how to translate keywords into their math equivalents. This way, you can understand exactly what a problem is asking you to find and how you’re supposed to find it.

There are free SAT Math questions available online , so memorize your terms and then practice on realistic SAT word problems to make sure you’ve got your definitions down and can apply them to the actual test.

#3: Underline and/or Write Out Important Information

The key to solving a word problem is to bring together all the key pieces of given information and put them in the right places. Make sure you write out all these givens on the diagram you’ve drawn (if the problem calls for a diagram) so that all your moving pieces are in order.

One of the best ways to keep all your pieces straight is to underline your key information in the problem, and then write them out yourself before you set up your equation. So take a moment to perform this step before you zero in on solving the question.

#4: Pay Close Attention to What's Being Asked

It can be infuriating to find yourself solving for the wrong variable or writing in your given values in the wrong places. And yet this is entirely too easy to do when working with math word problems.

Make sure you pay strict attention to exactly what you’re meant to be solving for and exactly what pieces of information go where. Are you looking for the area or the perimeter? The value of x, 2x, or y?

It’s always better to double-check what you’re supposed to find before you start than to realize two minutes down the line that you have to begin solving the problem all over again.

#5: Brush Up on Any Specific Math Topic You Feel Weak In

You're likely to see both a diagram/equation problem and a word problem for almost every SAT Math topic on the test. This is why there are so many different types of word problems and why you’ll need to know the ins and outs of every SAT Math topic in order to be able to solve a word problem about it.

For example, if you don’t know how to find an average given a set of numbers, you certainly won’t know how to solve a word problem that deals with averages!

Understand that solving an SAT Math word problem is a two-step process: it requires you to both understand how word problems work and to understand the math topic in question. If you have any areas of mathematical weakness, now's a good time to brush up on them—or else SAT word problems might be trickier than you were expecting!

body_ready-1

All set? Let's go!

Test Your SAT Math Word Problem Knowledge

Finally, it's time to test your word problem know-how against real SAT Math problems:

Word Problems

1. No Calculator

body_sat_math_test_question_1

2. Calculator OK

body_sat_math_test_question_2

3. Calculator OK

body_sat_math_test_question_3

4. Calculator OK

body_sat_math_test_question_4

Answers: C, B, A, 1160

Answer Explanations

1. For this problem, we have to use the information we're given to set up an equation.

We know that Ken spent x dollars, and Paul spent 1 dollar more than Ken did. Therefore, we can write the following equation for Paul:

Ken and Paul split the bill evenly. This means that we'll have to solve for the total amount of both their sandwiches and then divide it by 2. Since Ken's sandwich cost x dollars and Paul's cost x + 1, here's what our equation looks like when we combine the two expressions:

Now, we can divide this expression by 2 to get the price each person paid:

(2 x + 1) / 2

But we're not finished yet. We know that both Ken and Paul also paid a 20% tip on their bills. As a result, we have to multiply the total cost of one bill by 0.2, and then add this tip to the bill. Algebraically, this looks like this:

( x + 0.5) + 0.2( x + 0.5)

x + 0.5 + 0.2 x + 0.1

1.2 x + 0.6

The correct answer choice is (C) 1.2 x + 0.6

2. You'll have to be familiar with statistics in order to understand what this question is asking.

Since Nick surveyed a random sample of his freshman class, we can say that this sample will accurately reflect the opinion (and thus the same percentages) as the entire freshman class.

Of the 90 freshmen sampled, 25.6% said that they wanted the Fall Festival held in October. All we have to do now is find this percentage of the entire freshmen class (which consists of 225 students) to determine how many total freshmen would prefer an October festival:

225 * 0.256 = 57.6

Since the question is asking "about how many students"—and since we obviously can't have a fraction of a person!—we'll have to round this number to the nearest answer choice available, which is 60, or answer choice (B).

3. This is one of those problems that is asking you to define a value in the equation given. It might look confusing, but don't be scared—it's actually not as difficult as it appears!

First off, we know that t represents the number of seconds passed after an object is launched upward. But what if no time has passed yet? This would mean that t = 0. Here's what happens to the equation when we plug in 0 for t :

h (0) = -16(0)2 + 110(0) + 72

h (0) = 0 + 0 + 72

As we can see, before the object is even launched, it has a height of 72 feet. This means that 72 must represent the initial height, in feet, of the object, or answer choice (A).

4. You might be tempted to draw a diagram for this problem since it's talking about a pool (rectangle), but it's actually quicker to just look at the numbers given and work from there.

We know that the pool currently holds 600 gallons of water and that water has been hosed into it at a rate of 8 gallons a minute for a total of 70 minutes.

To find the amount of water in the pool now, we'll have to first solve for the amount of water added to the pool by hose. We know that 8 gallons were added each minute for 70 minutes, so all we have to do is multiply 8 by 70:

8 * 70 = 560 gallons

This tells us that 560 gallons of water were added to our already-filled, 600-gallon pool. To find the total amount of water, then, we simply add these two numbers together:

560 + 600 = 1160

The correct answer is 1160.

body_sleepy-1

Aaaaaaaaaaand time for a nap.

Key Takeaways: Making Sense of SAT Math Word Problems

Word problems make up a significant portion of the SAT Math section, so it’s a good idea to understand how they work and how to translate the words on the page into a proper expression or equation. But this is still only half the battle.

Though you won’t know how to solve a word problem if you don’t know what a product is or how to draw a right triangle, you also won’t know how to solve a word problem about ratios if you don’t know how ratios work.

Therefore, be sure to learn not only how to approach math word problems as a whole, but also how to narrow your focus on any SAT Math topics you need help with. You can find links to all of our SAT Math topic guides here to help you in your studies.

What’s Next?

Want to brush up on SAT Math topics? Check out our individual math guides to get an overview of each and every topic on SAT Math . From polygons and slopes to probabilities and sequences , we've got you covered!

Running out of time on the SAT Math section? We have the know-how to help you beat the clock and maximize your score .

Been procrastinating on your SAT studying? Learn how you can overcome your desire to procrastinate and make a well-balanced prep plan.

Trying to get a perfect SAT score? Take a look at our guide to getting a perfect 800 on SAT Math , written by a perfect scorer.

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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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  • \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
  • \mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
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  • How do you solve word problems?
  • To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
  • How do you identify word problems in math?
  • Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
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  • Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
  • What is an age problem?
  • An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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SAT Math Word Problems

Most students tremble with fear when they think about preparing for the Math portion of the SAT .  After all, there are 20+ topics tested, and the test requires mastery of topics ranging from arithmetic to trigonometry. Some topics cover basic math questions from algebra class, such as solving linear equations or recognizing graphs of parabolas. Solving questions such as these is tough enough, but the SAT adds a layer of difficulty by presenting word problems from a wide range of disciplines that require you to use your algebra skills to solve them.

To tackle these word problems, you must first understand the context of the word problem and then create the equations that you’ll need to solve, based on the information provided in the question.

In this article, I’ll first cover two critical algebra skills you need to master. Then I’ll provide a number of strategies for efficiently solving hard SAT word problems, as well as examples, so we can see these strategies in action and avoid common mistakes.

Here are the topics we’ll cover:

Example 1: linear equation with one variable, sales tax: example 1, substitution method practice 1, substitution method practice 2, when dealing with word problems, translation is key, balancing equations: example 1, balancing equations: example 2, balancing equations: example 3, “number of items”: example 1, “number of items”: example 2, age problem: example 1, money problem: example 1, in conclusion: sat math word problems, what’s next.

We’ll start with solving basic linear equations, then we’ll review the substitution method of solving SAT algebra problems. Finally, we’ll use these two techniques to solve math word problem examples.

Basic Algebra – Solving Single-Variable Linear Equations 

The simplest algebra questions on the SAT are single-variable linear equations. Your goal is to solve for the value of the variable:

  • 2v + 5 = 3v – 7

Note that solving single-variable equations generally requires that we combine all terms containing the variable on one side of the equation and all constant terms on the other side.

Let’s practice simplifying and solving an equation with one variable.

If 8q – 2 = 10q + 14, then what is the value of q?

First, we combine our constant terms by adding 2 to both sides:

8q – 2 + 2 = 10q + 14 + 2

8q = 10q + 16

Next, we combine the variable terms by subtracting 10q from both sides of the equation:

8q – 10q = 10q + 16 – 10q

Finally, we divide both sides of the equation by -2 to isolate the variable:

-2q / -2 = 16 / -2

When solving for a single variable, we combine like terms and isolate the variable we are solving for.

Now, let’s practice with an example.

If -2x – 11 = 5x + 3, then x equals which of the following?

-2x – 11 = 5x + 3

-2x = 5x + 14

Now let’s turn our attention to applying algebra to solve a straightforward word problem involving sales tax.

Sales Tax Questions

A common single-variable word problem that you might be asked on the SAT involves the calculation of sales tax. When you buy an item, you not only pay for the item itself but you also generally pay a surcharge called sales tax, typically stated as a percentage of the cost of the item. We add the sales tax to the cost of the item to determine the actual amount we will pay at the register.

The equation for purchasing an item in an area where sales tax is charged is:

Cost of Item + Sales Tax = Total

Let’s look at a basic example of the calculation of sales tax.

Laura buys a blender for $40.00 in a town where the sales tax rate is 4.5%. How much will she have to pay, including sales tax?

We see that the cost of the item is $40.00, and the sales tax rate is 4.5%. Thus, the sales tax amount is 4.5% of $40.00, or 40 x 0.045. Let’s put these into the equation:

40 + (40)(0.045) = Total

40 + 1.8 = Total

41.8 = Total

Thus, Laura will pay a total of $41.80: the item cost of $40.00 and the sales tax of $1.80.

Know the sales tax formula: Cost of Item + Sales Tax = Total

Let’s look at a sample SAT question about sales tax.

Alex buys a pair of shoes. The sales tax rate in his town is 6%. The register total for his purchase, including sales tax, is $115.54. How much sales tax did Alex pay?

We set up the linear equation by noting that the total purchase price is equal to the cost of the item plus the sales tax:

Total = Cost of Item + Sales Tax

We are given the total and the sales tax rate. However, we don’t know the actual cost of the item, so we will let x = the cost of the item:

115.54 = x + (0.06)(x)

We combine like terms on the right side of the equation and solve for x:

115.54 = 1.06x

115.54 / 1.06 = x

We see that the cost of the item is $109.00. Now we calculate the sales tax:

109 x 0.06 = 6.54

The sales tax is $6.54.

Next, let’s discuss linear equations with two variables.

Linear Equations with Two Variables/Systems of Linear Equations

Two-variable equations have not just one but two different variables:

  • v – u = 12
  • 2x + y = 10
  • 5z – 3y = 42

These are called two-variable equations because they contain two different variables . If we have a second equation that contains one or both of the variables, the two equations are called a system of linear equations. One common way to solve for the values of the variables is by using the substitution method . Let’s discuss the substitution method now.

Using the Substitution Method to Solve a System of Linear Equations

When we’re working with SAT systems of equations word problems, the essence of the substitution method is that we first isolate one variable in one of the equations. Then we substitute whatever that variable is equal to into the other equation. Let’s practice with an example.

In the system of linear equations below, determine the value of a.

b = 3a (equation 1)

a + b = 12 (equation 2)

Looking at our two equations, we see that b is already isolated in equation 1. So, we can substitute 3a (from equation one) for b (in equation two). This gives us:

a + 3a = 12

For the system of linear equations below, what is the value of x?

3x + y = 11 (equation 1)

2x + 5y = 3 (equation 2)

First, let’s isolate y in equation 1 by subtracting 3x from both sides of the equation:

y = 11 – 3x

Now, we can substitute 11 – 3x for y in equation 2. This gives us:

2x + 5(11 – 3x) = 3

2x + 55 – 15x = 3

We can use the substitution method to solve a system of linear equations.

Next, let’s discuss the many topics on the SAT in which we use algebra to determine the solution.

Here are a couple of sample SAT math word problems that you might encounter:

Luca buys apples, which cost $2 each, and bananas, which cost $3 each, at the market. If he buys twice as many bananas as apples and spends $48 at the market, how many bananas does he buy?

Tyrone and Greg have a total of 40 marbles. If Tyrone has 4 times the number of marbles that Greg has, how many marbles does Greg have?

The bottom line is that a word problem presents a scenario requiring us to translate the given information into an algebraic equation that we then solve.

Word problems are not just about solving equations; they are also about translating words into equations! Let’s look at some common translations:

  • “Is” translates to equals ( = )

Kendra is the same age as Carla

Kendra’s age = Carla’s age

  • “More” translates to addition ( + )

Arita has 6 more marbles than Pablo

Arita = Pablo + 6

  • “Less/fewer” translates to subtraction (-)

Sammy has 3 fewer coins than Rati

Sammy = Rati – 3

  • “Times as many” translates to multiplication (✖)

Harold has 5 times as many newspapers as Phoebe

Harold = Phoebe ✖ 5

Know the common translations of words to algebraic equations.

Before jumping into word problem practice questions, let’s discuss one point of confusion students have when translating words into equations: properly balancing the equations.

You Must Make Sure Your Equations are Balanced

A basic principle for translating words into equations is that the equations must be balanced correctly. If they are not, you might obtain reversed values for your variables’ values, or you might get a negative answer. Let’s look at a few correct and incorrect ways to balance equations, and we’ll explain why the equations are balanced correctly.

Sherry has 30 more dollars than Melanie.

When we are setting up this equation, we must understand that Sherry has more money than Melanie. For example, if Melanie has 20 dollars, then Sherry has 50 dollars.

So, to properly balance the equation, we must add 30 dollars to Melanie’s amount and set it equal to Sherry’s amount.

If we let S = Sherry’s money and M = Melanie’s money, we have:

Nala has 4 fewer toys than Frank.

When considering this equation, we must understand that Nala has fewer toys than Frank.

In other words, Frank has more toys than Nala.

So, to set up a balanced equation, we must subtract 4 from Frank’s amount and set that equal to Nala’s.

If we let N = the number of Nala’s toys and F = the number of Frank’s toys, we have:

N – 4 = F

There are 5 times as many baseballs as tennis balls.

When considering this equation, we must understand that there are more baseballs than tennis balls.

So, to set up a correct equation, we multiply the number of tennis balls by 5 and set that amount equal to the number of baseballs.

If we let B = the number of baseballs and T = the number of tennis balls, we have:

Now that we are familiar with how to translate, balance, and solve equations, let’s jump into the various types of general word problems you will encounter on the SAT, and how to solve them.

“Number of Items” Questions

We have already encountered some simple examples of “number of items” questions in this article. Now, let’s attack a full-fledged problem like those you might encounter on test day.

Julia and Tory have a total of 30 candy bars. If Julia has 5 times as many candy bars as Tory, how many candy bars does Tory have?

First, we define our variables.

Let’s let J = the number of candy bars Julia has and T = the number of candy bars Tory has.

Next, let’s create equations based on the information given in the question stem.

“Julia and Tory have a total of 30 candy bars” is translated as:

J + T = 30 (equation 1)

“Julia has 5 times as many candy bars as Tory” is translated as:

J = 5T (equation 2)

Next, we use the substitution method by substituting 5T for J in equation 1:

5T + T = 30

Let’s practice one more. This time, we will add a twist.

Leti has 4 times as many cookies as Henrik. If Leti gives Henrik 7 cookies, Henrik will have 4 fewer cookies than Leti. How many cookies did Leti originally have?

We can let L = the number of cookies Leti has and H = the number of cookies Henrik has. Next, we use the information from the question stem to create equations.

Because Leti has 4 times as many cookies as Henrik, we have:

L = 4H (equation 1)

After Leti gives Henrik 7 cookies, Leti will have L – 7 cookies. After Henrik gets the 7 cookies from Leti, he will have H + 7 cookies. He will still have 4 fewer cookies than Leti.

Thus, our second equation is:

L – 7 = (H + 7) +4

L = H + 18 (equation 2)

Next, we can use the substitution method, substituting 4H for L in equation 2:

4H = H + 18

Henrik originally had 6 cookies. Thus, Leti originally had 4 x 6 = 24 cookies.

Next, let’s discuss another common type of SAT word problem: age problems.

Age Problems

Age problems are a common type of SAT problem with a unique spin. Instead of having a straightforward translation, as we practiced above, an age problem will usually make us translate words into equations based on an age in the past, an age in the future, or even both.

For example, we might need to consider Ann’s age 10 years from now. To do this, we let Ann’s age today = A, so her age in 10 years will be A + 10.

Similarly, Ann’s age 12 years ago would be expressed as A – 12.

Age problems generally require us to compare ages in the past or future.

Chantal is 6 years younger than Yolanda. If, in 5 years, Yolanda will be twice as old as Chantal, how old is Chantal today?

Let’s let C = Chantal’s age today and Y = Yolanda’s age today. Next, let’s create some equations.

Because Chantal is 6 years younger than Yolanda, we have:

C = Y – 6 (equation 1)

Now we express what each age will be in 5 years:Chantal will be C + 5 and Yolanda will be Y + 6. The relationship between their ages in 5 years is that Yolanda will be twice Chantal’s age:

Y + 5 = 2(C + 5)

Y + 5 = 2C + 10 (equation 2)

Next, we use the substitution method and substitute Y – 6 for C in equation 2, and we have:

Y + 5 = 2(Y – 6) + 10

Y + 5 = 2Y – 12 + 10

Y + 5 = 2Y – 2

Yolanda is 7 years old today. Thus, Chantal is 7 – 6 = 1 year old today.

Next, let’s discuss money problems.

Money Problems

In general, money problems on the SAT involve two different commodities, such as adult tickets and child tickets, and the revenue earned from selling them. We can create two equations, one relating the number of items and the second relating the revenue earned from the sale of the items.

We will see that the substitution method is extremely useful for solving the two equations that are created from the given information.

Be prepared to use the substitution method when dealing with money problems.

At an ice cream stand, customers can buy either a small cone for $3.00 or a large cone for $4.50. Yesterday, the owner reported he had sold 720 cones, and his revenue for the day was $2,340. How many small cones did he sell?

First, let’s define our variables. We can let the number of small cones sold = x and the number of large cones sold = y.

Next, let’s create our equations. Because the owner sold 720 ice cream cones, we have:

x + y = 720 (equation 1)

His total revenue was $2,340 from selling x small cones for $3.00 each and y large cones for $4.50 each:

3x + 4.5y = 2,340 (equation 2)

Let’s isolate x in equation 1:

x = 720 – y

Now, let’s substitute 720 – y for x into equation 2 and solve:

3(720 – y) + 4.5y = 2,340

2,160 – 3y + 4.5y = 2,340

Since the number of large cones sold is 120, the number of small cones sold is 720 – 120 = 600.

Of the 20+ major math topics on the SAT, one of the most challenging is word problems. In order to be successful with this type of problem, you need to be skilled both at translating words into equations and at solving algebra equations. The ability to solve SAT word problems is a must for scoring well on the exam.

The two most common skills for solving math word problems are knowing how to solve a linear equation and using the substitution method for solving systems of linear equations. You must translate the question into an equation or equations first, then use the appropriate algebraic method to arrive at the answer.

Some of the common types of word problems that we have covered here are sales tax questions, “number of items” questions, age questions, and money questions.

In this article, we have covered how to solve word problems on the SAT. You have a good start in doing well on this particular topic. But don’t lose your perspective: word problems are only one of many topics you must master in order to get a great SAT score.  Read our article on how to improve your SAT score for more strategies, tips, and tricks.

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COMMENTS

  1. Algebraic word problems

    Algebraic word problems are questions that require translating sentences to equations, then solving those equations. The equations we need to write will only involve basic arithmetic operations and a single variable. Usually, the variable represents an unknown quantity in a real-life scenario.

  2. Equation Word Problems Worksheets

    This compilation of a meticulously drafted equation word problems worksheets is designed to get students to write and solve a variety of one-step, two-step and multi-step equations that involve integers, fractions, and decimals. These worksheets are best suited for students in grade 6 through high school.

  3. 1.20: Word Problems for Linear Equations

    Solution: Adding 9 to a number is written as x + 9, while subtracting 7 from three times the number is written as 3x − 7. We therefore get the equation: x + 9 = 3x − 7. We solve for x by adding 7 on both sides of the equation: x + 16 = 3x. Then we subtract x: 16 = 2x. After dividing by 2, we obtain the answer x = 8.

  4. Solving Word Questions

    Solving Word Questions With LOTS of examples! In Algebra we often have word questions like: Example: Sam and Alex play tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play? How do we solve them? The trick is to break the solution into two parts: Turn the English into Algebra.

  5. How to write word problems as equations

    What is the problem actually asking you to do? There are certain phrases that always mean the same operation in math. The table below will help you learn common phrases in math and what operations they represent.

  6. Solving Equations: Word Problems

    DEFINE a variable. Figure out what the unknown in the problem is and assign a variable to it. For example it could be, Let x=the number or Let h=the \# of hot dogs. WRITE an equation. Check out how to translate words into expressions. SOLVE the equation. CHECK your answer. STATE your answer.

  7. Equation Word Problems

    12 Like Alex Federspiel Video 4 (Video) Writing Equations from Word Problems by Katie Tatum Here's a video by Katie Tatum that shows you how to create equations from a word problem. Summary This is a very important and useful skill to learn how to do. The trick is to locate key words that signal you to what needs to be done.

  8. Algebra Topics: Introduction to Word Problems

    12 - 4 12 - 4 = 8, so you know Johnny has 8 apples left. Word problems in algebra If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

  9. Solving Algebra Word Problems

    Step 1: Analyze the problem. Step 2: Gather information. Step 3: Translate into an equation. Step 4: Solve the equation. What are examples of word problems? Example 1: Harry and David...

  10. 120 Math Word Problems To Challenge Students Grades 1 to 8

    The list of examples is supplemented by tips to create engaging and challenging math word problems. 120 Math word problems, categorized by skill Addition word problems Best for: 1st grade, 2nd grade 1. Adding to 10: Ariel was playing basketball. 1 of her shots went in the hoop. 2 of her shots did not go in the hoop.

  11. Algebra Word Problems

    Identify the variable: Use the statement, Let x = _____. You can replace the x with whatever variable you are using. Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem. The following key words will help you write equations for Algebra word problems: Addition altogether

  12. 7.3 Simple Algebraic Equations and Word Problems

    Step 5: After completing Step 4, there should be a single variable with a coefficient of +1 on the left side and a single constant term on the right side - that constant term is the solution to the equation. Step 6: Verify the answer by substituting the solution from Step 5 back into the original problem. Step 7:

  13. Solving Equations

    Example 1: solve equations involving like terms. Solve for x. x. 5q-4q=9 5q −4q = 9. Combine like terms. Combine the q q terms on the left side of the equation. To do this, subtract 4q 4q from both sides. (5 q-4 q)=9-4 q (5q −4q) = 9− 4q. The goal is to simplify the equation by combining like terms.

  14. Solving Systems of Equations Word Problems

    Example 1: Systems Word Problems You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined.

  15. Solving Linear Equation Word Problems (9 Terrific Examples!)

    Solving equations and word problems Example But, what about the tricks and steps? Yes, there are some easy to follow steps that we are going to use to solve linear word problems. Read the problem carefully and determine what is being asked. Create a sidebar!

  16. Polynomial Equation Word Problems (video lessons, examples and solutions)

    Learn to write and solve polynomial equations for special integers, consecutive integers. Example 1: Find a number that is 56 less than its square. Let n be the number. Example 2: Find two consecutive odd integers whose sum is 130. Try the free Mathway calculator and problem solver below to practice various math topics.

  17. Solving Word Problems: Steps & Examples

    Let me show you a 3-step process you can use to help you solve your word problems. You will need your algebra skills for solving equations with variables in them, so access that part of your brain ...

  18. 6.8 Mixture and Solution Word Problems

    6.8 Mixture and Solution Word Problems. Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below: Example ...

  19. How to Solve System of Equations Word Problems

    A word problem can be translated into a system of equations. Example Problem: A piggy bank of coins has quarters and dimes totaling $2.90. If there are 8 more dimes than quarters, how many of each ...

  20. The Complete Guide to SAT Math Word Problems

    Word Problem Type 1: Setting Up an Equation. This is a fairly uncommon type of SAT word problem, but you'll generally see it at least once on the Math section. You'll also most likely see it first on the section. For these problems, you must use the information you're given and then set up the equation.

  21. Word Problems Calculator

    Examples Frequently Asked Questions (FAQ) How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform.

  22. SAT Math Word Problems

    In Conclusion: SAT Math Word Problems. Of the 20+ major math topics on the SAT, one of the most challenging is word problems. In order to be successful with this type of problem, you need to be skilled both at translating words into equations and at solving algebra equations. The ability to solve SAT word problems is a must for scoring well on ...