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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?}
  • \mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
  • \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
  • \mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
  • 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.
  • Is there a calculator that can solve word problems?
  • 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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  • High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,... Read More

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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?

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Mathematics LibreTexts

1.20: Word Problems for Linear Equations

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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?

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How to Solve Word Problems in Algebra

Last Updated: December 19, 2022 Fact Checked

This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 68,958 times.

You can solve many real world problems with the help of math. In order to familiarize students with these kinds of problems, teachers include word problems in their math curriculum. However, word problems can present a real challenge if you don't know how to break them down and find the numbers underneath the story. Solving word problems is an art of transforming the words and sentences into mathematical expressions and then applying conventional algebraic techniques to solve the problem.

Assessing the Problem

Step 1 Read the problem carefully.

  • For example, you might have the following problem: Jane went to a book shop and bought a book. While at the store Jane found a second interesting book and bought it for $80. The price of the second book was $10 less than three times the price of he first book. What was the price of the first book?
  • In this problem, you are asked to find the price of the first book Jane purchased.

Step 3 Summarize what you know, and what you need to know.

  • For example, you know that Jane bought two books. You know that the second book was $80. You also know that the second book cost $10 less than 3 times the price of the first book. You don't know the price of the first book.

Step 4 Assign variables to the unknown quantities.

  • Multiplication keywords include times, of, and f actor. [9] X Research source
  • Division keywords include per, out of, and percent. [10] X Research source
  • Addition keywords include some, more, and together. [11] X Research source
  • Subtraction keywords include difference, fewer, and decreased. [12] X Research source

Finding the Solution

Step 1 Write an equation.

Completing a Sample Problem

Step 1 Solve the following problem.

  • Robyn and Billy run a lemonade stand. They are giving all the money that they make to a cat shelter. They will combine their profits from selling lemonade with their tips. They sell cups of lemonade for 75 cents. Their mom and dad have agreed to double whatever amount they receive in tips. Write an equation that describes the amount of money Robyn and Billy will give to the shelter.

Step 2 Read the problem carefully and determine what you are asked to find.

  • Since you are combining their profits and tips, you will be adding two terms. So, x = __ + __.

.75c

Expert Q&A

Daron Cam

Video . By using this service, some information may be shared with YouTube.

  • Word problems can have more than one unknown and more the one variable. Thanks Helpful 2 Not Helpful 1
  • The number of variables is always equal to the number of unknowns. Thanks Helpful 1 Not Helpful 0
  • While solving word problems you should always read every sentence carefully and try to extract all the numerical information. Thanks Helpful 1 Not Helpful 0

how to solve equation word problems

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Calculate Frequency

  • ↑ Daron Cam. Academic Tutor. Expert Interview. 29 May 2020.
  • ↑ http://www.purplemath.com/modules/translat.htm
  • ↑ https://www.mathsisfun.com/algebra/word-questions-solving.html
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut8_probsol.htm
  • ↑ http://www.virtualnerd.com/algebra-1/algebra-foundations/word-problem-equation-writing.php
  • ↑ https://www.khanacademy.org/test-prep/praxis-math/praxis-math-lessons/praxis-math-algebra/a/gtp--praxis-math--article--algebraic-word-problems--lesson

About This Article

Daron Cam

To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Next, summarize what information you know and what you need to know. Then, assign variables to the unknown quantities. For example, if you know that Jane bought 2 books, and the second book cost $80, which was $10 less than 3 times the price of the first book, assign x to the price of the 1st book. Use this information to write your equation, which is 80 = 3x - 10. To learn how to solve an equation with multiple variables, keep reading! Did this summary help you? Yes No

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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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I wrote ____ in the place of total marbles since that is what the problem is asking for (the unknown).

All of this may look oversimplified, but helping children to see the underlying relationship between the quantities is important. Consider now this problem:

Example: Jenny and Kenny together have 37 marbles, and Kenny has 15. How many does Jenny have? Many teachers might try to explain this as a subtraction problem, but in the most fundamental level it is about addition! It still talks about two people having certain amount of marbles together . The relationship between the quantities is the SAME as above, so we still need to write an addition equation. Relationship: Jenny's marbles  +  Kenny's marbles  =  Total marbles Equation: _____ + 15 = 37 Then, we can solve the equation ____ + 15 = 37  by subtracting. Using this kind of approach in the elementary grades will help children to set up equations in algebra story problems later.
Example : Jenny, Kenny, and Penny together have 51 marbles. Kenny has double as many marbles as Jenny has, and Penny has 12. How many does Jenny have? The relationship between the quantities is the same, so it is solved the same way: by writing an addition equation. However, we need to denote the number of Jenny's and Kenny's marbles with something. Jenny's marbles are unknown, so we can denote that with the variable n . Then Kenny has 2 n marbles. Relationship: Jenny's marbles  +  Kenny's marbles  +  Penny's marbles  =  Total marbles Equation: n + 2 n  +  12 = 51
Example: Jane is on page 79 of her book. The book has 254 pages. How many pages does she still have to read? This time the word " still " clues us in to an additive relationship where one of the addends is missing. You can initially write an empty line for what is not known, and later replace that with a variable. pages already read  +  pages still to read  =  pages total + = This equation is of course is then solved by subtraction, but it is better if you view it as an addition situation and write an addition equation for it.
Example:   The number of hours that were left in the day was one-third of the number of hours already passed. How many hours were left in the day? (From Grade 5 word problems for kids ) Can you see the general principle governing this problem?  It talks about the hours of the day where some hours already passed and some hours left. This, of course, points to addition once again: we have one part of the day, another part, and a total. The only quantity we know is the total hours for the day. We don't know the hours already passed nor the hours left, so initially you can use two empty lines in the equation that shows the basic relationship between the quantities: hours already passed  +  hours left = total hours = Then, the information in the first sentence gives us another relationship: "The number of hours that were left in the day was one-third of the number of hours already passed." We don't know the amount of hours passed nor the hours left. So let's use the variable p for the hours passed. Then we can write an expression involving p for the hours left, because "hours left is one-third of the hours passed," or hours left  =  1/3 p Then writing 1/3 p for the "hours left" in the first equation will give us: hours already passed + hours left = total hours p + 1/3 p = 24 This can be solved using basic algebra or by guess & check.

Subtraction word problems

One situation that indicates subtraction is difference or  how many/much more . However, the presence of the word "more" can indicate either an addition or subtraction, so be careful.

Example:   Ted read 17 pages today, and Fred read 28. How many more pages did Fred read? The solution is of course 28 − 17 = 11, but it's not enough to simply announce that – children need also to understand that difference is the result of a subtraction and tells the answer to how many more . Relationship:    Pages Fred read  −  pages Ted read = difference Equation: 28  −  17 = __ Example:   Greg has 17 more marbles than Jack. Jack has 15. How many does Greg have? Here the word more has a different meaning. This problem is not about the difference. The question asks how many does Greg have – not what is the difference in the amounts of marbles. It simply states Greg has 17 more compared to Jack, so here the word more simply indicates addition: Greg has as many as Jack AND 17 more, so Greg has 15 + 17 marbles.
Example: The mass of the Great Pyramid is 557t greater than that of the Leaning Tower of Pisa. Stone Henge has a mass of 2695t which is 95t less than the Leaning Tower of Pisa. There once was a Greater Pyramid which had a mass twice that of the Great Pyramid. What was the mass of the Greater Pyramid? (From Grade 5 word problems for kids ) Each of the first three sentences give information that can be translated into an equation. The question is not about how many more so it's not about difference. One thing being greater than another implies you add. One thing being less than another implies you subtract. And one thing being twice something indicates multiplying by 2. When I read this problem, I could immediately see that I could write equations from the different sentences in the problem, but I couldn't see the answer right away. I figured that after writing the equations I would see some way forwad; probably one equation is solved and gives an answer to another equation. The first sentence says, "The mass of the Great Pyramid is 557t greater than that of the Leaning Tower of Pisa". What are the quantities and the relationship between them here? mass of Great Pyramid = mass of the Leaning Tower of Pisa + 557t The second sentence says "Stonehenge has a mass of 2695t which is 95t less than the Leaning Tower of Pisa."  Here it gives you a relationship similar to the one above, and it actually spells out the mass of Stonehenge. It's like two separate pieces of information: "Stonehenge weighs 95t less than the tower.  Stonehenge weighs 2695t."  Less means you subtract. If you have trouble deciding which is subtracted from which, you can think in your mind which is heavier: Stonehenge or the tower?   either      mass of Stonehenge = mass of tower − 95t or mass of tower = mass of Stonehenge − 95t Now since the mass of Stonehenge is given, you can solve this equation, and from that knowledge you can solve the first equation, and from that go on to the mass of the " Greater Pyramid ".

If the teacher just jumps directly to the number sentences when solving word problems, then the students won't see the step that happens in the mind before that. The quantities and the relationship between them have to be made clear and written down before fiddling with the actual numbers. Finding this relationship should be the most important part of the word problems.  One could even omit the actual calculations and concentrate just finding the quantities and relationships.

Problem of Helen's hair length

Problem.   Helen has 2 inches of hair cut off each time she goes to the hair salon. If h equals the length of hair before she cuts it and c equals the length of hair after she cuts it, which equation would you use to find the length of Helen's hair after she visit the hair salon? a. h = 2 − c      c. c = h − 2 b. c = 2 − h      d. h = c − 2

Solution.   Ignoring the letters c and h for now, what are the quantities?  What principle or relationship is there between them? Which possibility of the ones listed below is right?  Which do you take away from which?

SIMPLE, isn't it??  In the original problem, the equations are given with the help of h and c instead of the long phrases "hair length before cutting" and "hair length after cutting". You can substitute the c , h , and 2 into the relationships above, and then match the equations (1) - (4) with the equations (a) through (d).

Helping students to write the algebraic equations

One idea that came to mind is to go through the examples above, and more, based on the typical word problems in the math books, and then turn the whole thing around and have students do exercises such as:

money earned − money spent on this - money spent on that = money left

original price − discount percent x original price = discounted price

money earned each month − expenses/taxes each month = money to use each month AND money to use each month × number of months = money to use over a period of time

speed × time = distance AND distance from A to B + distance from B to C = distance from A to C

Why are math word problems SO difficult for elementary school children? Hint: it has to do with a "recipe" that many math lessons follow.

The do's and don'ts of teaching problem solving in math General advice on how you can teach problem solving in elementary, middle, and high school math.

How I Teach Word Problems by Andre Toom (PDF) This article is written by a Russian who immigrated to US and noticed how COLLEGE LEVEL students have difficulties even with the simplest word problems! He describes his ideas on how to fill in the gap formed when students haven't learned how to solve word problems in earlier education.

A list of websites focusing on word problems and problem solving Use these sites to find good word problems to solve. Most are free!

When solving word problems, students must first decide what quantity represents x and then must write all the other quuantities in terms of x. I teach the students to set up arrows according to the language in the problem. All arrows point to x. Example. Harry had 10 less toys than Mark. Sue has twice as many toys as Harry. Set up arrows: Sue--- Harry---Mark Therefore Mark is x, Harry is x-10 and Sue is 2(x-10). Students find this very helpful. Sandy Denny
My idea is the math teacher might teach and understand the students at the same time and everyone would have a sense of humor. So I think she/he will know if the students are listening or not, when after the class, talk to the student and ask what is wrong. Don't hurt the student's feelings. lorence

Practice makes perfect. Practice math at IXL.com

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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.

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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:

how to solve equation word problems

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
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Mathx

  • Two-step equations – word problems

Simply put, two-step equations – word problems are two step equations expressed using words instead of just numbers and mathematical symbols. They are just a bit more complicated than one-step equations with word problems and they demand just a bit more effort to solve. If you are not confident in your abilities to solve two-step equations with word problems, you can go to one-step equations – word problems and practice some more before continuing with this lesson. But if you feel ready, we will show you how to solve it using this example:

Hermione’s Bikes rents bikes for $10 plus $4per hour. Janice paid $30 to rent a bike. For how many hours did she rent the bike?

First thing we have to do in this assignment is to find the variable and see what its connection is with the other values. The thing we do not know is the number of hours Janice rented the bike for and we have been asked to find that out. That means the number of hours is our variable.

The cost of renting a bike is 10$ to take the bike and 4$ for every hour it spends in our possession. The final sum Janice paid is $30. Let us write that down as an equation.

4 * x + 10 = 30

Now, in order to make things neater and more clear, let us move all the numbers (except for the number 4 – we have to get rid of it in a different way) to the right side of the equation. Like this:

4 * x = 30 – 10

To simplify things further, let us perform the subtraction.

The next thing to do is to get rid of the number 4 in front of the variable. We will do that by dividing the whole equation by 4.

4 * x = 20 |:4

Now that we have calculated the value of the variable, we can tell that Janice rented that bike for 5 hours. If you want to check the result – you can. If you multiply $4 that Janice paid per hour by the 5 hours she spend with that bike and then add the $10 she had to pay regardless of the time she spent with the bike, you will get a total sum of $30 that is indeed the full sum she paid.

two step equations word problems

These word problems are called two-step because you have to perform two mathematical operations in order to solve them. In this case – addition (subtraction) and multiplication (division). To practice solving two-step equations – word problems, feel free to use the worksheets below.

Two-step equations – word problems exams for teachers

Two-step equations – word problems worksheets for students.

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Linear Equation Word Problems - Examples & Practice - Expii

Linear equation word problems - examples & practice, explanations (3), (video) slope-intercept word problems.

by cpfaffinator

how to solve equation word problems

This video by cpfaffinator explains how to translate word problems into variables while working through a couple of examples.

(Note: the answer at 3:34 should be y=0.50x+3 instead of y=0.5x+b .)

When writing linear equations (also called linear models) from word problems, you need to know what the x and y variables refer to, as well as what the slope and y-intercept are. Here are some tricks to help translate :

  • The x variable. Usually, you'll be asked to find an equation in terms of some factor. This will usually be whatever affects the output of the equation.
  • The y variable. The y variable is usually whatever you're trying to find in the problem. It'll depend on the x variable.
  • The slope (m). The slope will be some sort of rate or other change over time.
  • The y-intercept (b). Look for a starting point or extra fees—something that will be added (or subtracted) on , no matter what the x variable is.

Using these parts, we can write the word problem in slope intercept form , y=mx+b, and solve.

A city parking garage charges a flat rate of $3.00 for parking 2 hours or less, and $0.50 per hour for each additional hour. Write a linear model that gives the total charge in terms of additional hours parked.

First up, we want the equation, or model, to give us the total charge. This will be what the y variable represents. y=total charge for parking The total charge depends on how many additional hours you've parked. This is the x variables. x=additional hours parked

Next, we need to find the slope, m. If we look at the slope-intercept equation , y=mx+b, the slope is multiplied with the input x. So, we want to find some rate in our problem that goes with the number of additional hours parked . In the problem, we have $0.50 per hour for each additional hour. The number of hours is being multiplied by the price of 50 cents every hour. m=0.50

The y-intercept is the starting point. In other words, if we parked our car in the garage for 0 hours, what would the price be? The problem says the garage *charges a flat rate of $3.00. A flat rate means that no matter what we're being charged 3 dollars. b=3

Plugging these into the slope-intercept equation will give us our model. y=mx+by=0.50x+3

Six 2 foot tall pine trees were planted during the school's observation of Earth Awareness Week in 1990. The trees have grown at an average rate of 34 foot per year. Write a linear model that gives the height of the trees in terms of the number of years since they were planted.

What does the y variable represent in our equation?

Years since the trees were planted

Number of trees planted

Radius of trees' roots

Height of the trees

Related Lessons

Translate to a linear equation.

Sometimes, the trick to solving a word problem will be to translate it into a linear equation. To clue you in, linear equation word problems usually involve some sort of rate of change , or steady increase (or decrease) based on a single variable. If you see the word rate , or even "per" or "each" , it's a safe bet that a word problem is calling for a linear equation.

There are a couple steps when translating from a word problem to a linear equation. Review them, then we'll work through an example.

  • Find the y variable, or output. What is the thing you're trying to find? This will often be a price, or an amount of time, or something else countable that depends on other things.
  • Find the x variable, or input. What is affecting the price, or amount of time, etc.?
  • Find the slope. What's the rate in the problem?
  • Find the y-intercept. Is there anything that's added or subtracted on top of the rate, no matter what what x is?
  • Plug all the numbers you know into y=mx+b !
Wally and Cobb are starting a catering business. They rent a kitchen for $350 a month, and charge $75 for each event they cater. If they cater 12 events in a month, how much do they profit?

First, let's find the y variable. What are we trying to find in this problem?

The name of the business

how to solve equation word problems

Word Problems with Linear Relationships

When you're faced with a linear word problem, it can feel daunting to work out a solution. The good news is that there are few simple steps that you can take to tackle linear word problems.

how to solve equation word problems

Image source: by Anusha Rahman

For future linear word problems, you can use these same steps to tackle the problem!

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Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 × b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

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How to Solve Word Problems by Finding Two-Variable Equations?

In this step-by-step guide, you will learn how to solve word problems by finding two-variable equations.

How to Solve Word Problems by Finding Two-Variable Equations?

A step-by-step guide to solving word problems by finding two-variable equations

Here is a step-by-step guide on how to solve word problems by finding two-variable equations:

  • Read the problem carefully: Begin by reading the problem carefully to understand what is being asked. Identify the relevant variables and their relationships to each other.
  • Choose variables to represent the problem: Choose variables to represent the quantities involved in the problem. It’s common to use x and y as variables in two-variable equations, but other variables can also be used.
  • Identify the known quantities: Identify the quantities in the problem that are known, such as given values or constraints.
  • Identify the unknown quantities: Identify the quantities in the problem that are unknown, which are typically the values you are trying to solve for.
  • Write a two-variable equation: Write a two-variable equation that relates the known and unknown quantities. This equation should represent the problem in a way that can be solved algebraically.
  • Solve the equation: Use algebraic techniques to solve the equation for the unknown quantity. This may involve rearranging the equation, factoring, or using formulas.
  • Check your answer: Once you have solved for the unknown quantity, check your answer by plugging it back into the original equation and verifying that it makes sense in the context of the problem.
  • Interpret the solution: Finally, interpret the solution in the context of the problem. Make sure the answer makes sense and is reasonable given the known constraints and quantities.

Solving Word Problems by Finding Two-Variable Equations – Example 1

Solve the relationships in the word problem. There are \(8\) cookies in a pack. Let \(p\) represent the number of packs and \(c\) represent the number of cookies. Find the number of cookies when \(p=2\).

Look for relationships between the number of packs and cookies. Find c by multiplying the number of packs by cookies. So, \(8×2=16\) \(c=16\)

Solving Word Problems by Finding Two-Variable Equations – Example 2

Solve the relationships in the word problem. David rides the taxi for \(10\) minutes every day. \(d\) represents the number of days and \(m\) represents the total number of minutes David spends in the taxi. After two weeks how much time will he spend in the taxi?

Look for relationships between the number of days and minutes. He spends 10 minutes every day. To know the total time after two weeks (14 days), multiply time by days. So, \(10×14=140\) \(t=140\)

by: Effortless Math Team about 12 months ago (category: Articles )

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How to Solve Word Problems Requiring Quadratic Equations

Last Updated: December 27, 2020

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 14,258 times.

Some word problems require quadratic equations in order to be solved. In this article, you will learn how to solve those types of problems. Once you get the hang of it, it will be very easy.

Quadratic Equations

Step 1 Know what kind of problem you're tackling.

  • For the real life scenarios, factoring method is better.
  • In geometric problems, it is good to use the quadratic formula.

Real Life Scenario

Step 1 Ask to yourself,

  • In this problem, it asks for Kenny's birthday.

Step 2 Decide your variables.

  • Since negative month does not exist, 3 is the only one that makes sense.
  • Because the problem asks for both the month and the date, the answer would be March 18th. (Use the value for the other variable that you found in step 3.)

Geometric Problems

Step 1 Identify if it's a geometric problem.

  • In the problem above, it asks you only for the height of the triangle.

Step 3 Decide your variables.

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Solving word problem chart

1. Understand the Problem by Paraphrasing

2. identify key information and variables, 3. translate words into mathematical symbols, 4. break down the problem into manageable parts, 5. draw diagrams or visual representations, 6. use estimation to predict answers, 7. apply logical reasoning for unknown variables, 8. leverage similar problems as templates, 9. check answers in the context of the problem, 10. reflect and learn from mistakes.

Have you ever observed the look of confusion on a student’s face when they encounter a math word problem ? It’s a common sight in classrooms worldwide, underscoring the need for effective strategies for solving math word problems . The main hurdle in solving math word problems is not just the math itself but understanding how to translate the words into mathematical equations that can be solved.

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Generic advice like “read the problem carefully” or “practice more” often falls short in addressing students’ specific difficulties with word problems. Students need targeted math word problem strategies that address the root of their struggles head-on. 

A Guide on Steps to Solving Word Problems: 10 Strategies 

One of the first steps in tackling a math word problem is to make sure your students understand what the problem is asking. Encourage them to paraphrase the problem in their own words. This means they rewrite the problem using simpler language or break it down into more digestible parts. Paraphrasing helps students grasp the concept and focus on the problem’s core elements without getting lost in the complex wording.

Original Problem: “If a farmer has 15 apples and gives away 8, how many does he have left?”

Paraphrased: “A farmer had some apples. He gave some away. Now, how many apples does he have?”

This paraphrasing helps students identify the main action (giving away apples) and what they need to find out (how many apples are left).

Play these subtraction word problem games in the classroom for free:

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Students often get overwhelmed by the details in word problems. Teach them to identify key information and variables essential for solving the problem. This includes numbers , operations ( addition , subtraction , multiplication , division ), and what the question is asking them to find. Highlighting or underlining can be very effective here. This visual differentiation can help students focus on what’s important, ignoring irrelevant details.

  • Encourage students to underline numbers and circle keywords that indicate operations (like ‘total’ for addition and ‘left’ for subtraction).
  • Teach them to write down what they’re solving for, such as “Find: Total apples left.”

Problem: “A classroom has 24 students. If 6 more students joined the class, how many students are there in total?”

Key Information:

  • Original number of students (24)
  • Students joined (6)
  • Looking for the total number of students

Here are some fun addition word problems that your students can play for free:

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The transition from the language of word problems to the language of mathematics is a critical skill. Teach your students to convert words into mathematical symbols and equations. This step is about recognizing keywords and phrases corresponding to mathematical operations and expressions .

Common Translations:

  • “Total,” “sum,” “combined” → Addition (+)
  • “Difference,” “less than,” “remain” → Subtraction (−)
  • “Times,” “product of” → Multiplication (×)
  • “Divided by,” “quotient of” → Division (÷)
  • “Equals” → Equals sign (=)

Problem: “If one book costs $5, how much would 4 books cost?”

Translation: The word “costs” indicates a multiplication operation because we find the total cost of multiple items. Therefore, the equation is 4 × 5 = $20

Complex math word problems can often overwhelm students. Incorporating math strategies for problem solving, such as teaching them to break down the problem into smaller, more manageable parts, is a powerful approach to overcome this challenge. This means looking at the problem step by step rather than simultaneously trying to solve it. Breaking it down helps students focus on one aspect of the problem at a time, making finding the solution more straightforward.

Problem: “John has twice as many apples as Sarah. If Sarah has 5 apples, how many apples do they have together?”

Steps to Break Down the Problem:

Find out how many apples John has: Since John has twice as many apples as Sarah, and Sarah has 5, John has 5 × 2 = 10

Calculate the total number of apples: Add Sarah’s apples to John’s to find the total,  5 + 10 = 15

By splitting the problem into two parts, students can solve it without getting confused by all the details at once.

Explore these fun multiplication word problem games:

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Diagrams and visual representations can be incredibly helpful for students, especially when dealing with spatial or quantity relationships in word problems. Encourage students to draw simple sketches or diagrams to represent the problem visually. This can include drawing bars for comparison, shapes for geometry problems, or even a simple distribution to better understand division or multiplication problems .

Problem: “A garden is 3 times as long as it is wide. If the width is 4 meters, how long is the garden?”

Visual Representation: Draw a rectangle and label the width as 4 meters. Then, sketch the length to represent it as three times the width visually, helping students see that the length is 4 × 3 = 12

Estimation is a valuable skill in solving math word problems, as it allows students to predict the answer’s ballpark figure before solving it precisely. Teaching students to use estimation can help them check their answers for reasonableness and avoid common mistakes.

Problem: “If a book costs $4.95 and you buy 3 books, approximately how much will you spend?”

Estimation Strategy: Round $4.95 to the nearest dollar ($5) and multiply by the number of books (3), so 5 × 3 = 15. Hence, the estimated total cost is about $15.

Estimation helps students understand whether their final answer is plausible, providing a quick way to check their work against a rough calculation.

Check out these fun estimation and prediction word problem worksheets that can be of great help:

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When students encounter problems with unknown variables, it’s crucial to introduce them to logical reasoning. This strategy involves using the information in the problem to deduce the value of unknown variables logically. One of the most effective strategies for solving math word problems is working backward from the desired outcome. This means starting with the result and thinking about the steps leading to that result, which can be particularly useful in algebraic problems.

Problem: “A number added to three times itself equals 32. What is the number?”

Working Backward:

Let the unknown number be x.

The equation based on the problem is  x + 3x = 32

Solve for x by simplifying the equation to 4x=32, then dividing by 4 to find x=8.

By working backward, students can more easily connect the dots between the unknown variable and the information provided.

Practicing problems of similar structure can help students recognize patterns and apply known strategies to new situations. Encourage them to leverage similar problems as templates, analyzing how a solved problem’s strategy can apply to a new one. Creating a personal “problem bank”—a collection of solved problems—can be a valuable reference tool, helping students see the commonalities between different problems and reinforcing the strategies that work.

Suppose students have solved a problem about dividing a set of items among a group of people. In that case, they can use that strategy when encountering a similar problem, even if it’s about dividing money or sharing work equally.

It’s essential for students to learn the habit of checking their answers within the context of the problem to ensure their solutions make sense. This step involves going back to the original problem statement after solving it to verify that the answer fits logically with the given information. Providing a checklist for this process can help students systematically review their answers.

Checklist for Reviewing Answers:

  • Re-read the problem: Ensure the question was understood correctly.
  • Compare with the original problem: Does the answer make sense given the scenario?
  • Use estimation: Does the precise answer align with an earlier estimation?
  • Substitute back: If applicable, plug the answer into the problem to see if it works.

Problem: “If you divide 24 apples among 4 children, how many apples does each child get?”

After solving, students should check that they understood the problem (dividing apples equally).

Their answer (6 apples per child) fits logically with the number of apples and children.

Their estimation aligns with the actual calculation.

Substituting back 4×6=24 confirms the answer is correct.

Teaching students to apply logical reasoning, leverage solved problems as templates, and check their answers in context equips them with a robust toolkit for tackling math word problems efficiently and effectively.

One of the most effective ways for students to improve their problem-solving skills is by reflecting on their errors, especially with math word problems. Using word problem worksheets is one of the most effective strategies for solving word problems, and practicing word problems as it fosters a more thoughtful and reflective approach to problem-solving

These worksheets can provide a variety of problems that challenge students in different ways, allowing them to encounter and work through common pitfalls in a controlled setting. After completing a worksheet, students can review their answers, identify any mistakes, and then reflect on them in their mistake journal. This practice reinforces mathematical concepts and improves their math problem solving strategies over time.

3 Additional Tips for Enhancing Word Problem-Solving Skills

Before we dive into the importance of reflecting on mistakes, here are a few impactful tips to enhance students’ word problem-solving skills further:

1. Utilize Online Word Problem Games

A word problem game

Incorporate online games that focus on math word problems into your teaching. These interactive platforms make learning fun and engaging, allowing students to practice in a dynamic environment. Games can offer instant feedback and adaptive challenges, catering to individual learning speeds and styles.

Here are some word problem games that you can use for free:

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2. Practice Regularly with Diverse Problems

Word problem worksheet

Consistent practice with a wide range of word problems helps students become familiar with different questions and mathematical concepts. This exposure is crucial for building confidence and proficiency.

Start Practicing Word Problems with these Printable Word Problem Worksheets:

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3. Encourage Group Work

Solving word problems in groups allows students to share strategies and learn from each other. A collaborative approach is one of the best strategies for solving math word problems that can unveil multiple methods for tackling the same problem, enriching students’ problem-solving toolkit.

Conclusion 

Mastering math word problems is a journey of small steps. Encourage your students to practice regularly, stay curious, and learn from their mistakes. These strategies for solving math word problems are stepping stones to turning challenges into achievements. Keep it simple, and watch your students grow their confidence and skills, one problem at a time.

Frequently Asked Questions (FAQs)

How can i help my students stay motivated when solving math word problems.

Encourage small victories and use engaging tools like online games to make practice fun and rewarding.

What's the best way to teach beginners word problems?

Begin with simple problems that integrate everyday scenarios to make the connection between math and real-life clear and relatable.

How often should students practice math word problems?

Regular, daily practice with various problems helps build confidence and problem-solving skills over time.

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Can basic mathematics be done in Microsoft Word?

Is there a function of MS Word that enables users to solve basic math problems, in this case addition or subtraction? I use its platform for a budget and of course I could just use a calculator but it would be more convenient if I could solve it all in one place.

For instance:

Sorry for the simplicity of this question. Wondering if MS Word had a functionality like this of some sort?

  • microsoft-word

Hennes's user avatar

  • 2 A word processor as a spreadsheet? Really? –  Ignacio Vazquez-Abrams Mar 19, 2012 at 7:09
  • 1 Yeah, I know, I'm old school. –  Christopher Chipps Dec 20, 2012 at 21:36

5 Answers 5

Yes, Word can do this (the Equation Editor is built in; the computation and graphing stuff is part of a free add-in from Microsoft). While Tom's suggestion of Wolfram Alpha is nice, it's not really usable without an internet connection. This will work offline, and can be almost as capable (graphing included!).

Note that these directions are for Word 2010, but the add-in mentioned supports Word 2007. Some tabs and buttons might be in slightly different locations or have slightly different names.

Install the Microsoft Mathematics Add-In for Word and OneNote .

Insert an equation using the Equation Editor. This can be done with Alt + = or by clicking the Equation button in the Symbols section on the Insert tab.

Screenshot of insert equation

You should now have this box.

Screenshot of empty equation box

Enter your equation. Make sure the box is selected in some way, clicking inside it is enough. Note that for addition use + , for subtraction use - , for division use / and for multiplication use * . If you want a nice division symbol, use \div . If you want a nice multiplication symbol (x), use \times (with a space after it). Just x inserts the pronumeral x, as used in algebra. This can also do much more advanced things, such as display (but not calculate) \theta , display \infty (infinity). It can also display and calculate things like \sqrt(4) , and sin(45) . It also supports order of operations, and can actually rearrange algebraic formulae quite nicely. There are many neat little tricks you can learn over time, or ask even on SuperUser about.

Screenshot of filled equation box

Again, make sure the equation box is selected in some way . Go to the Mathematics tab (after installing the add-in), click Compute , then Calculate .

Screenshot of clicking calculate

The free Microsoft Mathematics 4.0 program can also do the same thing, and is even more powerful. It also supports copying equations directly from Word's built in Equation Editor (again, the add-in only adds computation and graphing capabilities), and you can copy its output directly back into Word. I'm reasonably sure the underlying engine for the program and add-in is the same, actually. They both rely on .NET 3.5 SP1.

Community's user avatar

  • 2 This doesn't work for the latest version of Word (2019). –  IllustriousMagenta Oct 15, 2019 at 7:32
  • @Modelmat Consider mrseanong.com/video-blog/… . But there's no guarantees, as this add-on appears to be long discontinued. –  Bob Oct 15, 2019 at 7:42
  • During installation, it says it will only work with Microsoft Word 2007 and 2010 –  CreativiTimothy Feb 5, 2021 at 6:29
  • confirming that this doesn't seem to be natively in desktop Word 365 today, and the link to install is bad now; the solution for Formula field here , superuser.com/a/1129026/288047 , is working fine for the basic arithmetic I want –  Mike M Sep 11, 2023 at 12:42

It is very easy to compute simple arithmetic expressions in Word 2010 . You need to use a Formula field as described below:

  • Press [Ctrl][F9] to insert a blank field, a pair of braces { } .
  • Type the following: =6.75+12.65+27.35
  • Note that there should a blank space after the left brace { and before the right brace }.
  • Press [F9] to update the result.

If everything goes well, you should see 46.75.

Abdul Quadir's user avatar

  • still works today , Microsoft® Word for Microsoft 365 MSO (Version 2308 Build 16.0.16731.20182) 64-bit –  Mike M Sep 25, 2023 at 17:47

So, if you don't care about the answer being inserted into the document automatically and you just want to know the answer to a simple equation already in a Word document, I think this it the best out-of-the-box, offline, no equation box required, no addins required option: https://www.howtogeek.com/267279/how-to-perform-simple-calculations-in-microsoft-word/

When you need to do a quick calculation, normally you would think to use the Windows calculator. However, if you’re working in Microsoft Word, you can calculate simple equations typed into your document using Word’s not-so-obvious Calculate command. To use the Calculate command, we need to add it to the Quick Access Toolbar. To do this, click the down arrow button on the right side of the Quick Access Toolbar and select “More Commands” from the drop-down menu. Select “All Commands” from the “Choose commands from” drop-down list. In the list of commands on the left, scroll down to the “Calculate” command, select it, and then click “Add”. The Calculate command is added to the list on the right. Click “OK” to accept the change. Now you can type in and then select a simple equation (do not select the equals sign) in your Word document and click the “Formula” button. For some reason, the button is not called Calculate.

You do have to select/highlight the text you want it to evaluate, if you miss part of the selection then it doesn't get rolled into the answer. That wasn't quite explicitly clear in the instructions.

Word computation quick access equation and answer

If anyone knows the background of this magical button, or knows how to get it through a different means other than the quick access toolbar addition, that would be good to know.

Kalen Brown's user avatar

You can actually enter very simple formulae into tables in MS Word. Use the Table/Formula menu when you are in the table cell where you want the result. You can even use functions like average, count, product, sum. But it's really much, much easier to use a linked spreadsheet.

Lord Peter's user avatar

No, you can type exquations in Microsoft Word but you can't execute them.

What you want is provided in Microsoft Excel, at best you might be able to include a Excel Worksheet in your Microsoft Word document; however, I would think you'd rather want to use Microsoft Excel instead.

Or use Wolfram Alpha if you are making your homework in Word and need some place to copy your equations to to get solutions for them, it recognizes the complex equations quite well. Sometimes needs you to fix something that it misinterpreted...

Tamara Wijsman's user avatar

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how to solve equation word problems

IMAGES

  1. Solving Word Problems Using Two-Step Equations

    how to solve equation word problems

  2. How to Solve a Word Problem Using a Rational Equation

    how to solve equation word problems

  3. how to solve equations in word problems

    how to solve equation word problems

  4. How To Write Linear Equations From Word Problems

    how to solve equation word problems

  5. Printables

    how to solve equation word problems

  6. Using Algebra To Solve Word Problems

    how to solve equation word problems

VIDEO

  1. Linear Equation Word Problems

  2. Equation Word Problems Practice Sheet (Page 4.9) 2/1/23

  3. #DIFFERENTIAL EQUATION WORD PROBLEMS AND HOMOGENEOUS FUNCTIONS

  4. U3 Equation Word Problems

  5. Two-Step Equation

  6. Linear equation word problems

COMMENTS

  1. Algebraic word problems

    To solve an algebraic word problem: 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. It's important for us to keep in mind how we define our variables.

  2. Word Problems Calculator

    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.

  3. Solving Word Questions

    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!

  4. 1.20: Word Problems for Linear Equations

    Solution: Translating the problem into an algebraic equation gives: 2x − 5 = 13 2 x − 5 = 13. We solve this for x x. First, add 5 to both sides. 2x = 13 + 5, so that 2x = 18 2 x = 13 + 5, so that 2 x = 18. Dividing by 2 gives x = 182 = 9 x = 18 2 = 9. c) A number subtracted from 9 is equal to 2 times the number.

  5. How to Solve Word Problems in Algebra

    2. Solve an equation for one variable. If you have only one unknown in your word problem, isolate the variable in your equation and find which number it is equal to. Use the normal rules of algebra to isolate the variable. Remember that you need to keep the equation balanced.

  6. 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...

  7. Algebra Topics: Introduction to Word Problems

    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.

  8. Quadratic equations word problem

    Course: Algebra 1 > Unit 14 Lesson 10: Quadratic standard form Finding the vertex of a parabola in standard form Graphing quadratics: standard form Graph quadratics in standard form Quadratic word problem: ball Quadratic word problems (standard form) Math > Algebra 1 > Quadratic functions & equations > Quadratic standard form

  9. How to set up algebraic equations to match word problems

    original price − discount percent x original price = discounted price. Write 3 different story problems where the solution is based on the relationships. money earned each month − expenses/taxes each month = money to use each month AND. money to use each month × number of months = money to use over a period of time.

  10. 5 Tricks for Solving Algebra Word Problems

    How to Solve Algebra Problems: 5 Helpful Tips. When you solve algebra problems, it's smart to have a plan of attack ready to follow. Solving word problems may seem difficult, but when you read through the problem and can figure out what the specific equation is, it's no harder than a regular algebra problem.

  11. How to write word problems as equations

    How to write word problems as equations — Krista King Math | Online math help Word problems can seem to be tricky at first. 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.

  12. Solving word problems with Two-step Equations

    Learn how to solve word problems using two-step equations.For more Math help visit our websitehttp://www.moomoomath.com/

  13. How to Solve Systems of Equations Word Problems? (+FREE Worksheet!)

    Systems of Equations Word Problems - Example 1: Tickets to a movie cost \ ($8\) for adults and \ ($5\) for students. A group of friends purchased \ (20\) tickets for \ ($115.00\). How many adult tickets did they buy? Answer: Let \ (x\) be the number of adult tickets and \ (y\) be the number of student tickets. There are \ (20\) tickets.

  14. Solving Equations

    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. ... 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 ...

  15. Systems of equations word problems

    Math > Algebra 1 > Systems of equations > Systems of equations word problems Google Classroom You might need: Calculator Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h . If doubled, Malcolm's maximum speed would be 80 km/h more than Ravi's maximum speed. What were Malcolm's and Ravi's maximum speeds?

  16. Solving Systems of Equations Word Problems

    Solving Systems of Equations Word Problems 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.

  17. Algebra

    Yay Math in Studio presents a single word problem, the likes of which students around the world experience constantly. The joy for this problem is finding wa...

  18. Two-step equations

    Simply put, two-step equations - word problems are two step equations expressed using words instead of just numbers and mathematical symbols. They are just a bit more complicated than one-step equations with word problems and they demand just a bit more effort to solve.

  19. Linear Equation Word Problems

    Translate to a Linear Equation. Sometimes, the trick to solving a word problem will be to translate it into a linear equation. To clue you in, linear equation word problems usually involve some sort of rate of change, or steady increase (or decrease) based on a single variable.If you see the word rate, or even "per" or "each", it's a safe bet that a word problem is calling for a linear equation.

  20. Solving Word Problem Using Equations

    Solve the word problems by using equations. Here's how: Are those word problems not adding up? Solve the word problems by using equations. Here's how: SAT Math.

  21. Math Problem Solver

    Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

  22. How to Solve Word Problems by Finding Two-Variable Equations

    Solving Word Problems by Finding Two-Variable Equations - Example 2 Solve the relationships in the word problem. David rides the taxi for \(10\) minutes every day. \(d\) represents the number of days and \(m\) represents the total number of minutes David spends in the taxi.

  23. 3 Ways to Solve Word Problems Requiring Quadratic Equations

    Geometric Problems. 1. Identify if it's a geometric problem. Geometric problems that require quadratic equations are good to be solved using the quadratic formula because the answer could be irrational. The quadratic formula is Failed to parse (MathML if possible (experimental): Invalid response ("Math extension cannot connect to Restbase ...

  24. 10 Best strategies for solving math word problems in 2024

    2. Identify Key Information and Variables. Students often get overwhelmed by the details in word problems. Teach them to identify key information and variables essential for solving the problem. This includes numbers, operations (addition, subtraction, multiplication, division), and what the question is asking them to find.Highlighting or underlining can be very effective here.

  25. Can basic mathematics be done in Microsoft Word?

    5 Answers Sorted by: 3 Yes, Word can do this (the Equation Editor is built in; the computation and graphing stuff is part of a free add-in from Microsoft). While Tom's suggestion of Wolfram Alpha is nice, it's not really usable without an internet connection. This will work offline, and can be almost as capable (graphing included!).

  26. Writing an Equation

    3:20 Word Problem Equations; 4:29 Lesson Summary; ... Let's take a couple of moments to review what we've learned about equations and formulas and how to solve them. First, an equation is a way to ...

  27. Hoboken Public School District on Instagram: "2nd Grade at Connors

    3 likes, 0 comments - hobokenpublicschooldistrict on February 7, 2024: "2nd Grade at Connors School: Ms. Rinaldi, Ms. Garcia, Ms. Criqui, Ms. Courtney, Mr. Donovan ...