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Praxis Core Math

Course: praxis core math   >   unit 1.

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

Algebraic word problems | Lesson

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

What are algebraic word problems?

What skills are needed.

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

How do we solve algebraic word problems?

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

What's a Venn diagram?

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

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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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  • Middle School Math Solutions – Inequalities Calculator Next up in our Getting Started maths solutions series is help with another middle school algebra topic - solving... Read More

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Equation Word Problems Worksheets

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

One Step Equation Word Problem Worksheets

One Step Equation Word Problem Worksheets

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

(15 Worksheets)

Two-Step Equation Word Problems: Integers

Two-Step Equation Word Problems: Integers

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

  • Download the set

Multi-Step Equation Word Problems: Integers

Multi-Step Equation Word Problems: Integers

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

Two-Step Equation Word Problems: Fractions and Decimals

Two-Step Equation Word Problems: Fractions and Decimals

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

Multi-Step Equation Word Problems: Fractions and Decimals

Multi-Step Equation Word Problems: Fractions and Decimals

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

MCQ - Equation Word Problems

MCQ - Equation Word Problems

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

Related Worksheets

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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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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.

/en/algebra-topics/solving-equations/content/

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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equations to solve word problems

Word Problems Linear Equations

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

Solving Equations: Word Problems - Examples & Practice - Expii

Solving equations: word problems - examples & practice, explanations (4).

equations to solve word problems

(Videos) Steps to Solving Word Problems

by mathman1024

equations to solve word problems

This video by mathman1024 goes through the steps of solving word problems based on linear equations.

Let's be real for a second. Word problems are the worst. But sadly, you can't avoid them. Luckily there are tricks and steps you can learn to make solving them easier. It will still take lot's of practice but hopefully they can lead you in the right direction. The steps to solving word problems are:

  • READ and understand the problem. It may seem like common sense but it's very important to read the whole problem and understand what it's about.
  • DEFINE a variable. Figure out what the unknown in the problem is and assign a variable to it. For example it could be, Let x=the number or Let h=the \# of hot dogs.
  • WRITE an equation. Check out how to translate words into expressions .
  • SOLVE the equation.
  • CHECK your answer.
  • STATE your answer. It's always good practice to state your answer at the end of a word problem. If the problem asks, "How many hot dogs did Lilly eat?" instead of saying h=6 write "Lilly ate 6 hot dogs."

He then goes over an example:

Five times a number plus nineteen is eighty-four. What is the number?

Step 1: READ the problem. It's okay if you need to read it multiple times to fully understand.

Step 2: DEFINE a variable. The problem is asking you to find a number. So we can say, Let x=the number.

Step 3: WRITE an equation. We know that we are multiplying a number, x, by five, so we can start with 5x. We then add 19 to that number, which gives us 5x+19. This is all equal to 84, so our full equation is 5x+19=84.

Step 4: SOLVE the equation. 5x+19=845x+19−19=84−195x=655x5=655x=13

Remember: whatever you do to one side of the equation, you have to do to the other . You may need to use the Addition and Subtraction Property of Equality or the Multiplication and Division Property of Equality .

Step 5: CHECK the answer. Check by plugging in 13 everywhere we see x. 5x+19=845(13)+19=8465+19=8484=84 This is a true statement, so our answer is correct.

Step 6: STATE your answer. The number is 13.

(Video) Solving Word Problems

equations to solve word problems

This video works through a similar type of problem. We need to translate the problem

The sum of four times some number and forty-three is the same as three times the difference of the number and seventy-eight. Find the number.

We can use this statement to write an equation.

The left side of the equation is "the sum of four times some number and forty-three". We can make the unknown number x. Four times this number gives us 4x. The sum of that value and 43 is 4x+43.

The right side of the equation is "three times the difference of the number and seventy-eight". This implies that our unknown number is the same number as the left side of the equation, so we'll use x again. "Three times the distance of..." implies a use of distribution over parentheses with subtraction inside, so we can start with 3(x−0). The last part is "difference of the number and seventy-eight", so we can subtract 78 inside of the parentheses, 3(x−78).

Now we can set the two sides equal to each other and solve for x. 4x+43=3(x−78) First, we're going to distribute 3 over x−78. 4x+43=3x−234 Now, we can isolate x on the left side by subtracting 3x and 43 from both sides. 4x+43−3x=3x−234−3xx+43−43=234−43x=−277

Our final answer is x=−277.

Related Lessons

(videos) word problems.

equations to solve word problems

This video by mathman1024 works through a word problem based on the length of different boards. Problems like this get easier when you draw yourself a picture to work with.

The problem states:

A board that is 131 inches long is cut into two pieces. One piece is fourteen inches longer than twice the other. How long is each piece?

Sometimes it's helpful to draw a picture or diagram when working on word problems. Not every problem requires you to do so but in some, like this one, is can be beneficial. Let's draw our board before we set up the equation.

A board

Image source: by Alex Federspiel

Now it says we're cutting it into two pieces. But notice that it doesn't say two equal pieces. This clues us in that we shouldn't cut the board down the middle. So let's just cut it randomly to the side.

A board cut randomly.

We have two different pieces now but don't use two variables. Often in these problems you'll have one part you know nothing about and another part that's built off that. For example if you say, "Mira has \textdollar10 more than Sam," you don't know how much either has. But once you find out how much Sam has, you can find out how much Mira has.

In this problem, which piece do we know nothing about? The short piece. We know the long piece is fourteen inches longer than twice the other . Let's have our variable represent the length of the short piece. Let x=length of short piece

The smaller side of board is labeled x.

The other piece is fourteen inches longer than twice the other . So that means its length is 2x+14.

The longer side is labeled 2x+14.

Image source: By Alex Federspiel

This is a two-step equation word problem.

How do we relate these two pieces? Well the first sentence says a board that is 131 inches long . So we know the combined length of these two pieces equals 131 inches. In other words, short piece+long piece=total board x+2x+14=131 From here we can solve like any other equation. x+2x+14=1313x+14=1313x+14−14=131−143x=1173x3=1173x=39 Now x is the length of the short piece. But the problem asks for the length of both pieces. So we need to plug our x-value into the expression for the long piece. 2x+142(39)+1478+1492 So finally we state our answer. The short piece is 39 inches.The long piece is 92 inches.

equations to solve word problems

Here's another video with another word problem. This one is similar to the one above so try this yourself first.

The second angle of a triangle is four degrees less than the first angle, and the third angle is twice the first angle. What is the measure of each angle of the triangle?

We know that we have three angles in a triangle , so we can name them m1, m2, and m3. We can set these equal to values.

  • m1 is going to be our base angle, so m1=x.
  • m2 is 4 degrees less than m1, so m2=x−4.
  • m3 is double m1, so m3+2x.

We also know that the angles of a triangle need to add to 180∘, so x+x−4+2x=180.

We can solve this equation for x. First, we're going to combine like terms on the left side. x+x−4+2x=1804x−4=180

Now, we're going to add 4 to both sides. 4x−4=1804x−4+4=180+44x=184

Finally, we'll divide both sides by 4 to get a value for x. 4x=1844x4=1844x=46

Now we can plug our x value into the definitions for our angles. m1=xm1=46∘m2=x−4m2=46−4m2=42∘m3=2xm3=2(46)m3=92∘

The last step is to test our angles to make sure they add up to 180. 46+42+92=180

So we have angles of 46∘,42∘, and 92∘.

Word problems are math problems written in another language. In order to manage these problems, we need to take the words and translate them into mathematical expressions . Once we do that, we can solve these equations just like we normally would.

equations to solve word problems

Image source: by Anusha Rahman

Let's take a look at an example of one of these two-step word problems :

On Saturdays, Jackson cuts grass and washes cars. He charges $15 dollars per lawn and an additional fee for each car he washes. Last Saturday, Jackson mowed 5 lawns and washed 10 cars. He made a total of $90 How much does Jackson charge per car wash?

equations to solve word problems

Image source: By Clker-Free-Vector-Images CC0 via Pixabay

First, we need to figure out what the question is asking. What are we trying to find out?

What are we trying to find out in this word problem?

How many cars Jackson washed

How many lawns Jackson cut

How much Jackson charges per car

How much money Jackson made in total

equations to solve word problems

**Solving Word Problems **

To solve word problems, you need to translate the problem into numerical formulas. Follow these steps to solve word problems. 1. Identify the unknown value in the problem and assign it a variable. For example,  Let x= the number of dogs 2. Identify the known values (constants) in the equation. 3. Write an equation using the unknown and known variables. 4. Solve the equation.

You can check you answer by plugging it back into the unknown variable and solving. Let’s look at an example.

Jack bought 35 pencils at the beginning of the school semester. Throughout the semester, he lost some pencils. At the end of the semester, he had** 13 pencils left**. How many pencils did Jack lose?

Step 1 : Identify the unknown value in the problem and assign it a variable. The problem is asking you to find out how many pencils Jack lost. Assign this value x.

Step 2 : Identify the known values (constants) in the equation. We know that Jack initially had 35 pencils, and ended the semester with 13 pencils.

Step 3 : Write an equation using the unknown and known variables. We know that we started with 35 pencils, and since we lost x pencils, we know that we subtract x from 35. This will equal to 13, which is the final number of pencils that Jack is left with. So the equation would be: 35−x=13

Step 4 : Solve the equation.

35−x=1335−x−13=13−1335−13−x=022−x=022−x+x=0+x22=xx=22

Jack lost 22 pencils during the semester.

We can check our answer by plugging 22 back into the unknown variable x and solve. 35−x=1335−(22)=1335−22=1313=13

This statement is true, so our answer is correct.

Now, try an example by yourself.

Leah is taking y classes in school. Each class gives about 30 minutes, or 0.5 hours of homework. She also wants to leave some time to hang out with her friends for 2 hours. At the end of the day, she calculated that she spent 5.5 hours on schoolwork and hanging out with her friends. How many classes is Leah taking?

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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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  • 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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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:

equations to solve 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

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

Learn how to create and solve systems of equations word problems by the elimination method.

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

Related Topics

  • How to Solve One-Step Equations
  • How to Solve One-Step Inequalities
  • How to Solve Multi-Step Inequalities
  • How to Solve Systems of Equations
  • How to Graph Single–Variable Inequalities

Step by step guide to solve systems of equations word Problems

  • Find the key information in the word problem that can help you define the variables.
  • Define two variables: \(x\) and \(y\)
  • Write two equations.
  • Use the elimination method for solving systems of equations.
  • Check the solution by substituting the ordered pair into the original equations.

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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?

Let \(x\) be the number of adult tickets and \(y\) be the number of student tickets. There are \(20\) tickets. Then: \(x+y=20\). The cost of adults’ tickets is \($8\) and for students ticket is \($5\), and the total cost is \($115\). So, \(8x+5y=115\). Now, we have a system of equations: \(\begin{cases}x+y=20 \\ 8x+5y=115\end{cases}\) Multiply the first equation by \(-5\) and add to the second equation: \(-5(x+y= 20)=- \ 5x-5y=- \ 100\) \(8x+5y+(-5x-5y)=115-100→3x=15→x=5→5+y=20→y=15\). There are \(5\) adult tickets and \(15\) student tickets.

Exercises for Solving Systems of Equations Word Problems

  • A farmhouse shelters \(10\) animals, some are pigs and some are ducks. Altogether there are \(36\) legs. How many of each animal are there?
  • A class of \(195\) students went on a field trip. They took vehicles, some cars, and some buses. Find the number of cars and the number of buses they took if each car holds \(5\) students and each bus holds \(45\) students.
  • The difference of two numbers is \(6\). Their sum is \(14\). Find the numbers.
  • The sum of the digits of a certain two–digit number is \(7\). Reversing is increasing the number by \(9\). What is the number?
  • The difference of two numbers is \(18\). Their sum is \(66\). Find the numbers.

Download Systems of Equations Word Problems Worksheet

  • There are \(8\) pigs and \(2\) ducks.
  • There are \(3\) cars and \(4\) buses.
  • \(10\) and \(4\).
  • \(24\) and \(42\).

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2.3: Use a Problem Solving Strategy

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Learning Objectives

By the end of this section, you will be able to:

  • Use a problem solving strategy for word problems
  • Solve number word problems
  • Solve percent applications
  • Solve simple interest applications

Be Prepared 2.4

Before you get started, take this readiness quiz.

Translate “six less than twice x ” into an algebraic expression. If you missed this problem, review Example 1.8.

Be Prepared 2.5

Convert 4.5% to a decimal. If you missed this problem, review Example 1.40.

Be Prepared 2.6

Convert 0.6 to a percent. If you missed this problem, review Example 1.40.

Have you ever had any negative experiences in the past with word problems? When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. Realize that your negative experiences with word problems are in your past. To move forward you need to calm your fears and change your negative feelings.

Start with a fresh slate and begin to think positive thoughts. Repeating some of the following statements may be helpful to turn your thoughts positive. Thinking positive thoughts is a first step towards success.

  I think I can! I think I can!

  While word problems were hard in the past, I think I can try them now.

  I am better prepared now—I think I will begin to understand word problems.

  I am able to solve equations because I practiced many problems and I got help when I needed it—I can try that   with word problems.

  It may take time, but I can begin to solve word problems.

You are now well prepared and you are ready to succeed. If you take control and believe you can be successful, you will be able to master word problems.

Use a Problem Solving Strategy for Word Problems

Now that we can solve equations, we are ready to apply our new skills to word problems. We will develop a strategy we can use to solve any word problem successfully.

Example 2.14

Normal yearly snowfall at the local ski resort is 12 inches more than twice the amount it received last season. The normal yearly snowfall is 62 inches. What was the snowfall last season at the ski resort?

Try It 2.27

Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was three more than twice the number of notebooks. He bought seven textbooks. How many notebooks did he buy?

Try It 2.28

Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is eight more than twice the number of crossword puzzles. He completed 22 Sudoku puzzles. How many crossword puzzles did he do?

We summarize an effective strategy for problem solving.

Use a Problem Solving Strategy for word problems.

  • Step 1. Read the problem. Make sure all the words and ideas are understood.
  • Step 2. Identify what you are looking for.
  • Step 3. Name what you are looking for. Choose a variable to represent that quantity.
  • Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
  • Step 5. Solve the equation using proper algebra techniques.
  • Step 6. Check the answer in the problem to make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

Solve Number Word Problems

We will now apply the problem solving strategy to “number word problems.” Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy.

Example 2.15

The sum of seven times a number and eight is thirty-six. Find the number.

Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

Try It 2.29

The sum of four times a number and two is fourteen. Find the number.

Try It 2.30

The sum of three times a number and seven is twenty-five. Find the number.

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far, we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

Example 2.16

The sum of two numbers is negative fifteen. One number is nine less than the other. Find the numbers.

Try It 2.31

The sum of two numbers is negative twenty-three. One number is seven less than the other. Find the numbers.

Try It 2.32

The sum of two numbers is negative eighteen. One number is forty more than the other. Find the numbers.

Some number problems involve consecutive integers . Consecutive integers are integers that immediately follow each other. Examples of consecutive integers are:

1 , 2 , 3 , 4 −10 , −9 , −8 , −7 150 , 151 , 152 , 153 1 , 2 , 3 , 4 −10 , −9 , −8 , −7 150 , 151 , 152 , 153

Notice that each number is one more than the number preceding it. Therefore, if we define the first integer as n , the next consecutive integer is n + 1 . n + 1 . The one after that is one more than n + 1 , n + 1 , so it is n + 1 + 1 , n + 1 + 1 , which is n + 2 . n + 2 .

n 1 st integer n + 1 2 nd consecutive integer n + 2 3 rd consecutive integer etc. n 1 st integer n + 1 2 nd consecutive integer n + 2 3 rd consecutive integer etc.

We will use this notation to represent consecutive integers in the next example.

Example 2.17

Find three consecutive integers whose sum is −54 . −54 .

Try It 2.33

Find three consecutive integers whose sum is −96 . −96 .

Try It 2.34

Find three consecutive integers whose sum is −36 . −36 .

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers . Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

24, 26, 28 24, 26, 28

−12 , −10 , −8 −12 , −10 , −8

Notice each integer is two more than the number preceding it. If we call the first one n , then the next one is n + 2 . n + 2 . The one after that would be n + 2 + 2 n + 2 + 2 or n + 4 . n + 4 .

n 1 st even integer n + 2 2 nd consecutive even integer n + 4 3 rd consecutive even integer etc. n 1 st even integer n + 2 2 nd consecutive even integer n + 4 3 rd consecutive even integer etc.

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 63, 65, and 67.

63, 65, 67 63, 65, 67

n , n + 2 , n + 4 n , n + 2 , n + 4

n 1 st odd integer n + 2 2 nd consecutive odd integer n + 4 3 rd consecutive odd integer etc. n 1 st odd integer n + 2 2 nd consecutive odd integer n + 4 3 rd consecutive odd integer etc.

Does it seem strange to have to add two (an even number) to get the next odd number? Do we get an odd number or an even number when we add 2 to 3? to 11? to 47?

Whether the problem asks for consecutive even numbers or odd numbers, you do not have to do anything different. The pattern is still the same—to get to the next odd or the next even integer, add two.

Example 2.18

Find three consecutive even integers whose sum is 120 120 .

Try It 2.35

Find three consecutive even integers whose sum is 102.

Try It 2.36

Find three consecutive even integers whose sum is −24 . −24 .

When a number problem is in a real life context, we still use the same strategies that we used for the previous examples.

Example 2.19

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

Try It 2.37

According to the National Automobile Dealers Association, the average cost of a car in 2014 was $28,400. This was $1,600 less than six times the cost in 1975. What was the average cost of a car in 1975?

Try It 2.38

US Census data shows that the median price of new home in the U.S. in November 2014 was $280,900. This was $10,700 more than 14 times the price in November 1964. What was the median price of a new home in November 1964?

Solve Percent Applications

There are several methods to solve percent equations. In algebra, it is easiest if we just translate English sentences into algebraic equations and then solve the equations. Be sure to change the given percent to a decimal before you use it in the equation.

Example 2.20

Translate and solve:

ⓐ What number is 45% of 84? ⓑ 8.5% of what amount is $4.76? ⓒ 168 is what percent of 112?

Try It 2.39

Translate and solve: ⓐ What number is 45% of 80? ⓑ 7.5% of what amount is $1.95? ⓒ 110 is what percent of 88?

Try It 2.40

Translate and solve: ⓐ What number is 55% of 60? ⓑ 8.5% of what amount is $3.06? ⓐ 126 is what percent of 72?

Now that we have a problem solving strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes sense—since many of the applications we will solve involve everyday situations, you can rely on your own experience.

Example 2.21

The label on Audrey’s yogurt said that one serving provided 12 grams of protein, which is 24% of the recommended daily amount. What is the total recommended daily amount of protein?

Try It 2.41

One serving of wheat square cereal has 7 grams of fiber, which is 28% of the recommended daily amount. What is the total recommended daily amount of fiber?

Try It 2.42

One serving of rice cereal has 190 mg of sodium, which is 8% of the recommended daily amount. What is the total recommended daily amount of sodium?

Remember to put the answer in the form requested. In the next example we are looking for the percent.

Example 2.22

Veronica is planning to make muffins from a mix. The package says each muffin will be 240 calories and 60 calories will be from fat. What percent of the total calories is from fat?

Try It 2.43

Mitzi received some gourmet brownies as a gift. The wrapper said each brownie was 480 calories, and had 240 calories of fat. What percent of the total calories in each brownie comes from fat? Round the answer to the nearest whole percent.

Try It 2.44

The mix Ricardo plans to use to make brownies says that each brownie will be 190 calories, and 76 calories are from fat. What percent of the total calories are from fat? Round the answer to the nearest whole percent.

It is often important in many fields—business, sciences, pop culture—to talk about how much an amount has increased or decreased over a certain period of time. This increase or decrease is generally expressed as a percent and called the percent change .

To find the percent change, first we find the amount of change, by finding the difference of the new amount and the original amount. Then we find what percent the amount of change is of the original amount.

Find percent change.

  • Step 1. Find the amount of change. change = new amount − original amount change = new amount − original amount
  • Step 2. Find what percent the amount of change is of the original amount. change is what percent of the original amount? change is what percent of the original amount?

Example 2.23

Recently, the California governor proposed raising community college fees from $36 a unit to $46 a unit. Find the percent change. (Round to the nearest tenth of a percent.)

Try It 2.45

Find the percent change. (Round to the nearest tenth of a percent.) In 2011, the IRS increased the deductible mileage cost to 55.5 cents from 51 cents.

Try It 2.46

Find the percent change. (Round to the nearest tenth of a percent.) In 1995, the standard bus fare in Chicago was $1.50. In 2008, the standard bus fare was 2.25.

Applications of discount and mark-up are very common in retail settings.

When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate , usually given as a percent, is used to determine the amount of the discount. To determine the amount of discount , we multiply the discount rate by the original price.

The price a retailer pays for an item is called the original cost . The retailer then adds a mark-up to the original cost to get the list price , the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.

amount of discount = discount rate · original price sale price = original amount – discount price amount of discount = discount rate · original price sale price = original amount – discount price

The sale price should always be less than the original price.

amount of mark-up = mark-up rate · original price list price = original cost + mark-up amount of mark-up = mark-up rate · original price list price = original cost + mark-up

The list price should always be more than the original cost.

Example 2.24

Liam’s art gallery bought a painting at an original cost of $750. Liam marked the price up 40%. Find ⓐ the amount of mark-up and ⓑ the list price of the painting.

ⓐ  

Try It 2.47

Find ⓐ the amount of mark-up and ⓑ the list price: Jim’s music store bought a guitar at original cost $1,200. Jim marked the price up 50%.

Try It 2.48

Find ⓐ the amount of mark-up and ⓑ the list price: The Auto Resale Store bought Pablo’s Toyota for $8,500. They marked the price up 35%.

Solve Simple Interest Applications

Interest is a part of our daily lives. From the interest earned on our savings to the interest we pay on a car loan or credit card debt, we all have some experience with interest in our lives.

The amount of money you initially deposit into a bank is called the principal , P , and the bank pays you interest , I. When you take out a loan, you pay interest on the amount you borrow, also called the principal.

In either case, the interest is computed as a certain percent of the principal, called the rate of interest , r . The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable t , (for time) represents the number of years the money is saved or borrowed.

Interest is calculated as simple interest or compound interest. Here we will use simple interest.

Simple Interest

If an amount of money, P , called the principal, is invested or borrowed for a period of t years at an annual interest rate r , the amount of interest, I , earned or paid is

I = interest I = P r t where P = principal r = rate t = time I = interest I = P r t where P = principal r = rate t = time

Interest earned or paid according to this formula is called simple interest .

The formula we use to calculate interest is I = P r t . I = P r t . To use the formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

Example 2.25

Areli invested a principal of $950 in her bank account that earned simple interest at an interest rate of 3%. How much interest did she earn in five years?

I = ? P = $ 950 r = 3 % t = 5 years I = ? P = $ 950 r = 3 % t = 5 years

Try It 2.49

Nathaly deposited $12,500 in her bank account where it will earn 4% simple interest. How much interest will Nathaly earn in five years?

Try It 2.50

Susana invested a principal of $36,000 in her bank account that earned simple interest at an interest rate of 6.5 % . 6.5 % . How much interest did she earn in three years?

There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we do not know the rate.

Example 2.26

Hang borrowed $7,500 from her parents to pay her tuition. In five years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of simple interest?

I = $ 1500 P = $ 7500 r = ? t = 5 years I = $ 1500 P = $ 7500 r = ? t = 5 years

Try It 2.51

Jim lent his sister $5,000 to help her buy a house. In three years, she paid him the $5,000, plus $900 interest. What was the rate of simple interest?

Try It 2.52

Loren lent his brother $3,000 to help him buy a car. In four years, his brother paid him back the $3,000 plus $660 in interest. What was the rate of simple interest?

In the next example, we are asked to find the principal—the amount borrowed.

Example 2.27

Sean’s new car loan statement said he would pay $4,866,25 in interest from a simple interest rate of 8.5% over five years. How much did he borrow to buy his new car?

I = 4,866.25 P = ? r = 8.5 % t = 5 years I = 4,866.25 P = ? r = 8.5 % t = 5 years

Try It 2.53

Eduardo noticed that his new car loan papers stated that with a 7.5% simple interest rate, he would pay $6,596.25 in interest over five years. How much did he borrow to pay for his car?

Try It 2.54

In five years, Gloria’s bank account earned $2,400 interest at 5% simple interest. How much had she deposited in the account?

Access this online resource for additional instruction and practice with using a problem solving strategy.

  • Begining Arithmetic Problems

Section 2.2 Exercises

Practice makes perfect.

List five positive thoughts you can say to yourself that will help you approach word problems with a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often.

List five negative thoughts that you have said to yourself in the past that will hinder your progress on word problems. You may want to write each one on a small piece of paper and rip it up to symbolically destroy the negative thoughts.

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question.

There are 16 girls in a school club. The number of girls is four more than twice the number of boys. Find the number of boys.

There are 18 Cub Scouts in Troop 645. The number of scouts is three more than five times the number of adult leaders. Find the number of adult leaders.

Huong is organizing paperback and hardback books for her club’s used book sale. The number of paperbacks is 12 less than three times the number of hardbacks. Huong had 162 paperbacks. How many hardback books were there?

Jeff is lining up children’s and adult bicycles at the bike shop where he works. The number of children’s bicycles is nine less than three times the number of adult bicycles. There are 42 adult bicycles. How many children’s bicycles are there?

In the following exercises, solve each number word problem.

The difference of a number and 12 is three. Find the number.

The difference of a number and eight is four. Find the number.

The sum of three times a number and eight is 23. Find the number.

The sum of twice a number and six is 14. Find the number.

The difference of twice a number and seven is 17. Find the number.

The difference of four times a number and seven is 21. Find the number.

Three times the sum of a number and nine is 12. Find the number.

Six times the sum of a number and eight is 30. Find the number.

One number is six more than the other. Their sum is 42. Find the numbers.

One number is five more than the other. Their sum is 33. Find the numbers.

The sum of two numbers is 20. One number is four less than the other. Find the numbers.

The sum of two numbers is 27. One number is seven less than the other. Find the numbers.

One number is 14 less than another. If their sum is increased by seven, the result is 85. Find the numbers.

One number is 11 less than another. If their sum is increased by eight, the result is 71. Find the numbers.

The sum of two numbers is 14. One number is two less than three times the other. Find the numbers.

The sum of two numbers is zero. One number is nine less than twice the other. Find the numbers.

The sum of two consecutive integers is 77. Find the integers.

The sum of two consecutive integers is 89. Find the integers.

The sum of three consecutive integers is 78. Find the integers.

The sum of three consecutive integers is 60. Find the integers.

Find three consecutive integers whose sum is −3 . −3 .

Find three consecutive even integers whose sum is 258.

Find three consecutive even integers whose sum is 222.

Find three consecutive odd integers whose sum is −213 . −213 .

Find three consecutive odd integers whose sum is −267 . −267 .

Philip pays $1,620 in rent every month. This amount is $120 more than twice what his brother Paul pays for rent. How much does Paul pay for rent?

Marc just bought an SUV for $54,000. This is $7,400 less than twice what his wife paid for her car last year. How much did his wife pay for her car?

Laurie has $46,000 invested in stocks and bonds. The amount invested in stocks is $8,000 less than three times the amount invested in bonds. How much does Laurie have invested in bonds?

Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more than three times the amount she earned from her job at the college. How much did she earn from her job at the college?

In the following exercises, translate and solve.

ⓐ What number is 45% of 120? ⓑ 81 is 75% of what number? ⓐ What percent of 260 is 78?

ⓐ What number is 65% of 100? ⓑ 93 is 75% of what number? ⓐ What percent of 215 is 86?

ⓐ 250% of 65 is what number? ⓑ 8.2% of what amount is $2.87? ⓐ 30 is what percent of 20?

ⓐ 150% of 90 is what number? ⓑ 6.4% of what amount is $2.88? ⓐ 50 is what percent of 40?

In the following exercises, solve.

Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. Geneva wants to leave 16% of the total bill as a tip. How much should the tip be?

When Hiro and his co-workers had lunch at a restaurant near their work, the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?

One serving of oatmeal has 8 grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily amount of fiber?

One serving of trail mix has 67 grams of carbohydrates, which is 22% of the recommended daily amount. What is the total recommended daily amount of carbohydrates?

A bacon cheeseburger at a popular fast food restaurant contains 2070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?

A grilled chicken salad at a popular fast food restaurant contains 650 milligrams (mg) of sodium, which is 27% of the recommended daily amount. What is the total recommended daily amount of sodium?

The nutrition fact sheet at a fast food restaurant says the fish sandwich has 380 calories, and 171 calories are from fat. What percent of the total calories is from fat?

The nutrition fact sheet at a fast food restaurant says a small portion of chicken nuggets has 190 calories, and 114 calories are from fat. What percent of the total calories is from fat?

Emma gets paid $3,000 per month. She pays $750 a month for rent. What percent of her monthly pay goes to rent?

Dimple gets paid $3,200 per month. She pays $960 a month for rent. What percent of her monthly pay goes to rent?

Tamanika received a raise in her hourly pay, from $15.50 to $17.36. Find the percent change.

Ayodele received a raise in her hourly pay, from $24.50 to $25.48. Find the percent change.

Annual student fees at the University of California rose from about $4,000 in 2000 to about $12,000 in 2010. Find the percent change.

The price of a share of one stock rose from $12.50 to $50. Find the percent change.

A grocery store reduced the price of a loaf of bread from $2.80 to $2.73. Find the percent change.

The price of a share of one stock fell from $8.75 to $8.54. Find the percent change.

Hernando’s salary was $49,500 last year. This year his salary was cut to $44,055. Find the percent change.

In ten years, the population of Detroit fell from 950,000 to about 712,500. Find the percent change.

In the following exercises, find ⓐ the amount of discount and ⓑ the sale price.

Janelle bought a beach chair on sale at 60% off. The original price was $44.95.

Errol bought a skateboard helmet on sale at 40% off. The original price was $49.95.

In the following exercises, find ⓐ the amount of discount and ⓑ the discount rate (Round to the nearest tenth of a percent if needed.)

Larry and Donna bought a sofa at the sale price of $1,344. The original price of the sofa was $1,920.

Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.

In the following exercises, find ⓐ the amount of the mark-up and ⓑ the list price.

Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%. What was the list price of the bracelet?

Regina bought a handmade quilt at original cost $120 to sell in her quilt store. She marked the price up 55%. What was the list price of the quilt?

Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up. What price did he charge his customers for the tomatoes?

Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up. What price did she charge her customers for the roses?

Casey deposited $1,450 in a bank account that earned simple interest at an interest rate of 4%. How much interest was earned in two years?

Terrence deposited $5,720 in a bank account that earned simple interest at an interest rate of 6%. How much interest was earned in four years?

Robin deposited $31,000 in a bank account that earned simple interest at an interest rate of 5.2%. How much interest was earned in three years?

Carleen deposited $16,400 in a bank account that earned simple interest at an interest rate of 3.9% How much interest was earned in eight years?

Hilaria borrowed $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000, plus $1,200 interest. What was the rate of simple interest?

Kenneth lent his niece $1,200 to buy a computer. Two years later, she paid him back the $1,200, plus $96 interest. What was the rate of simple interest?

Lebron lent his daughter $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000, plus $3,000 interest. What was the rate of simple interest?

Pablo borrowed $50,000 to start a business. Three years later, he repaid the $50,000, plus $9,375 interest. What was the rate of simple interest?

In 10 years, a bank account that paid 5.25% simple interest earned $18,375 interest. What was the principal of the account?

In 25 years, a bond that paid 4.75% simple interest earned $2,375 interest. What was the principal of the bond?

Joshua’s computer loan statement said he would pay $1,244.34 in simple interest for a three-year loan at 12.4%. How much did Joshua borrow to buy the computer?

Margaret’s car loan statement said she would pay $7,683.20 in simple interest for a five-year loan at 9.8%. How much did Margaret borrow to buy the car?

Everyday Math

Tipping At the campus coffee cart, a medium coffee costs $1.65. MaryAnne brings $2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?

Tipping Four friends went out to lunch and the bill came to $53.75 They decided to add enough tip to make a total of $64, so that they could easily split the bill evenly among themselves. What percent tip did they leave?

Writing Exercises

What has been your past experience solving word problems? Where do you see yourself moving forward?

Without solving the problem “44 is 80% of what number” think about what the solution might be. Should it be a number that is greater than 44 or less than 44? Explain your reasoning.

After returning from vacation, Alex said he should have packed 50% fewer shorts and 200% more shirts. Explain what Alex meant.

Because of road construction in one city, commuters were advised to plan that their Monday morning commute would take 150% of their usual commuting time. Explain what this means.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

This table has four columns and five rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was use a problem-solving strategy for word problems. In row 3, the I can was solve number problems. In row 4, the I can was solve percent applications. In row 5, the I can was solve simple interest applications.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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In order to access this I need to be confident with:

Solving equations

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

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

What is solving an equation?

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

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

For example,

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

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

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

What is solving an equation?

Common Core State Standards

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

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

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

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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 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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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.

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    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.

  7. Algebra Topics: Introduction to Word Problems

    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?

  8. 1.20: Word Problems for Linear Equations

    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 5 x = 60. We solve this for x x : x = 60 5 = 12 x = 60 5 = 12. b) If 5 is subtracted from twice an unknown number, the difference is 13.

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  10. Algebra Word Problems

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    To solve word problems, you need to translate the problem into numerical formulas. Follow these steps to solve word problems. 1. Identify the unknown value in the problem and assign it a variable. For example, Let x= the number of dogs 2. Identify the known values (constants) in the equation. 3. Write an equation using the unknown and known ...

  14. 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.

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    We also know that the highest grade added to the lowest grade is 138. So, (highest grade) + (lowest grade) = 142. In terms of our variable, Step 5: Solve the equation. Step 6: Answer the question in the problem. The problem asks us to find the lowest grade. We decided that l would be the number, so we have l = 48.

  17. Two-step equations

    14935. September 3, 2019. 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 word problems and they demand just a bit more effort to solve.

  18. Solving Systems of Equations Word Problems

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    Solve Number Word Problems. We will now apply the problem solving strategy to "number word problems." Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy.

  21. Solving Linear Equations Using Word Problems

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  23. 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.

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