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1.20: Word Problems for Linear Equations

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

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

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

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

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

\[7-3 x\nonumber\]

c) 8 less than a number.

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

\[x-8\nonumber\]

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

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

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

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

Simplifying this expression gives:

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

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

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

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

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

Example 18.2

Solve the following word problems:

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

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

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

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

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

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

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

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

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

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

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

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

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

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

Solution: We have the equation:

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

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

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

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

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

\[x + 9 = 3x − 7.\nonumber\]

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

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

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

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

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

Example 18.3

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

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

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

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

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

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

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

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

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

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

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

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

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

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

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

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

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

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

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

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

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

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

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

Distributing and collecting like terms give

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiplying both sides by 10 gives

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

Subtracting 64 from both sides gives

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

and dividing by 20 gives

Exit Problem

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

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REAL WORLD PROBLEMS: How to Write Equations Based on Algebra Word Problems

I know that you often sit in class and wonder, "Why am I forced to learn about equations, Algebra and variables?"

But... trust me, there are real situations where you will use your knowledge of Algebra and solving equations to solve a problem that is not school related. And... if you can't, you're going to wish that you remembered how.

It might be a time when you are trying to figure out how much you should get paid for a job, or even more important, if you were paid enough for a job that you've done. It could also be a time when you are trying to figure out if you were over charged for a bill.

This is important stuff - when it comes time to spend YOUR money - you are going to want to make sure that you are getting paid enough and not spending more than you have to.

Ok... let's put all this newly learned knowledge to work.

Click here if you need to review how to solve equations.

There are a few rules to remember when writing Algebra equations:

Writing Equations For Word Problems

subtraction

Multiplication.

double (2x)

triple (3x)

quadruple (4x)

divided into

Let's look at an example of an algebra word problem.

Example 1: Algebra Word Problems

Linda was selling tickets for the school play.  She sold 10 more adult tickets than children tickets and she sold twice as many senior tickets as children tickets.

As you can see, this problem is massive!  There are 5 questions to answer with many expressions to write. 

Algebra word problem solutions

A few notes about this problem

1. In this problem, the variable was defined for you.  Let x represent the number of children’s tickets sold tells what x stands for in this problem.  If this had not been done for you, you might have written it like this:

        Let x = the number of children’s tickets sold

2. For the first expression, I knew that 10 more adult tickets were sold.  Since more means add, my expression was x +10 .  Since the direction asked for an expression, I don’t need an equal sign.  An equation is written with an equal sign and an expression is without an equal sign.  At this point we don’t know the total number of tickets.

3. For the second expression, I knew that my key words, twice as many meant two times as many.  So my expression was 2x .

4.  We know that to find the total price we have to multiply the price of each ticket by the number of tickets.  Take note that since x + 10 is the quantity of adult tickets, you must put it in parentheses!  So, when you multiply by the price of $5 you have to distribute the 5.

5.  Once I solve for x, I know the number of children’s tickets and I can take my expressions that I wrote for #1 and substitute 50 for x to figure out how many adult and senior tickets were sold.

Where Can You Find More Algebra Word Problems to Practice?

Word problems are the most difficult type of problem to solve in math. So, where can you find quality word problems WITH a detailed solution?

The Algebra Class E-course provides a lot of practice with solving word problems for every unit! The best part is.... if you have trouble with these types of problems, you can always find a step-by-step solution to guide you through the process!

Click here for more information.

The next example shows how to identify a constant within a word problem.

Example 2 - Identifying a Constant

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20.

1. Write an equation that models this situation.

2. How many minutes were charged on this bill?

solving equations word problems

Notes For Example 2

The last example is a word problem that requires an equation with variables on both sides.

Example 3 - Equations with Variables on Both Sides

You have $60 and your sister has $120. You are saving $7 per week and your sister is saving $5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve.

solving equations word problems

Notes for Example 3

I'm hoping that these three examples will help you as you solve real world problems in Algebra!

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Algebra Topics  - Introduction to Word Problems

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

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Algebra Topics: Introduction to Word Problems

Lesson 9: introduction to word problems.

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What are word problems?

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

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

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

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

Word problems in algebra

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

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

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:

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:

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:

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:

f equals half of $8 plus $25

f = half of 8 + 25

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.

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.

1/2s = 29 - 25

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 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:

m plus three times m equals $280

m + three times m = 280

m + 3m = 280

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

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 .

70 + 3 ⋅ 70 = 280

70 + 210 = 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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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!)

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Step by step guide to solve systems of equations word Problems

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 adults ticket 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

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How to Solve Word Problems in Algebra

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This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been viewed 53,833 times.

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

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

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Math Lesson 9.2.5 - Equations for solving Word Problems

Welcome to our Math lesson on Equations for solving Word Problems , this is the fifth lesson of our suite of math lessons covering the topic of Word Problems Involving Equations , you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Using Other Types of Equations to Solve Word Problems

There are many types of equations besides the first - order equations with one variable. We will discuss a few examples where two other types of equations (second - order equations with one variable and first - order equations with two variables) are used in word problems. Let's present a short overview of these two types of equations.

Second - order equations with one variable

Second - order equations with one variable include equations that contain a single variable raised at the second power. In the next tutorials, we will see that the general form of these equations (otherwise known as 'quadratic equations') is

The left part of such equations is identical to the general form of quadratic expressions discussed in tutorial 6.4, where we explained how to factorise such expressions.

In this paragraph, however, we will consider only a simplified form of quadratic equations, where the coefficient b is 0. Therefore, the problems discussed here will require the use of the reduced quadratic equation

Let's consider an example to clarify this point.

The area of a square is 4 cm 2 more than that of a rectangle of dimensions 9 cm × 5 cm. What is the side length of the square?

The unknown in this problem is the side length of the square, which we denote by x . Since the area of a rectangle is the product of its dimensions and that of a square is the second power of the side length, we obtain the equation

Hence, the side length of the square is

(√49 can be 7 or - 7. However, we consider only the positive root, as the side length cannot be negative.)

First - order equations with two variables

First - order equations with two variables have the general form

where a and b are coefficients, c is a constant, while x and y are the two variables. For example, 3x - 2y - 6 = 0 is a first - order equation with two variables as it contains two variables x and y where both are in the first power (hence the name "first - order equation with two variables").

To solve first - order equations with two variables means finding the values of the two variables. However, since there are a lot of possible combinations that can give a true result, we must have some additional information about the relationship between the two variables. For example, in our example, we may take x = 4 and y = 3 as possible solutions, as 3 · 4 - 2 · 3 - 6 = 0 . Yet, we may also take x = 2 and y = 0 as possible solutions, as 3 · 2 - 2 · 0 - 6 = 0 . Both pairs of solutions are true. However, in a word problem, usually, there are only a couple of values that fit the description. Therefore, as said earlier, we must have some extra information about the relationship between the two variables. This extra information usually forms another first - order equation and we solve it as a system (we will explain systems of equations in the last two tutorials of this chapter). However, for now, it is sufficient to know that we express one of the variables in terms of the other variable and in this way, the original equation turns into a first - order equation with one variable, which we already know how to solve.

For example, if we have given that in a class there are 30 students where the boys vs girls ratio is B:G = 2:3 , we can write

This is a first - order equation (variables are written as B and G ) with two variables, as we can write it as

As we said earlier, it is impossible to find the correct value of the two variables only through this equation, as there are many combinations that fit the description. Therefore, we use the help of the other clue, which says that the total number of students in the class is 30. Hence, we have the other equation

available, which helps to express one of the variables (for example the girls G) in terms of the other variable in the first equation. Therefore, since G = 30 - B , we can substitute this expression in the first equation to obtain

Now, we have a first - order equation with one variable, which can be easily solved. Thus,

In this first - order equation with one variable we have a = 5 and b = -60 . Therefore, the root B that gives the number of boys in the class is

Hence, the number of girls in the class is

The perimeter of a rectangle is 12 cm more than that of an equilateral triangle with a 16 cm side length. Calculate the dimensions of the rectangle if its length is 2 cm more than its width.

We can express the rectangle's length by x and its width by y . Since the perimeter (the sum of all sides length) of a rectangle is calculated by the formula P = x + x + y + y = 2x + 2y , and given that an equilateral triangle has three equal sides, we have

This is a first - order equation with two variables. Therefore, we need some additional information to solve it. This additional information is already available. We know that length is 2 cm more than width, i.e. x = y + 2 . Therefore, we substitute x with its equivalent expression in the main equation to obtain

In this way, we obtained a first - order equation with one variable, where a = 2 and b = -28 . Therefore, we obtain for the variable y:

Hence, since x + y = 30 , we have

This too is a first - order equation with one variable, where a = 1 and b = -16 . Therefore, we obtain for the other variable x :

Hence, the length of the rectangle is 16 cm and the width is 14 cm.

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Solving Equations: Word Problems

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Basics on the topic Solving Equations: Word Problems

Real world problems are not usually written as an algebra equation, so we must translate a math story into an equation.

How do we do that? By carefully analyzing the information that is presented in the form of a word problem. Look for keywords to indicate the operations you should use to write an equation. For example, five more than x can be written as x + 5. The statement, ten times the unknown number, is written as 10x. What is the meaning of two times the sum of 5 and x? The word sum tells us to put 5 and x inside parentheses, the result is 2 ( 5 + x ).

After you have written the equation to describe your math story, solve the equation by following the steps you have learned for solving algebraic equations. Remember if there are multi steps, combine like terms then use inverse operations to isolate the variable to calculate the solution.

It’s always a good idea to check your work by substituting the solution back into the equation you wrote then check the solution against the math story. If the solution works for the equation but not for the story, recheck your original equation.

Explain steps to solve algebraic equations. CCSS.MATH.CONTENT.HSA.REI.A.1

Transcript Solving Equations: Word Problems

We're at the secret hideout of Dr. Evil, located on Volcano Island. He needs some gold to buy a present for his mommy's birthday.

Mommy Evil likes bling, so she wants a ring with as many diamonds as possible. Every jewel on the ring cost a measely 10 gold bars (obviously the typical currency for villans). But the problem is: he's a bit short on gold.

Word Problem Example

To determine how much of his stuff he needs to sell, we should convert the information from the word problem into an equation .

First, he lists one of his private islands on Evil-Bay and his asking price is 50 gold bars - quite a bargain!

Next, he'll sell 20 of his most evil robotic rocket crocs for 5 gold bars each. Hopefully they sell quickly. There's not much time until Mommy Evil's birthday.

He wonders how much bling he can put on the ring if each diamond costs 10 gold bars each. Let the number of diamonds equal x.

Translating the Word Problem and Writing the Equation

Let's write an equation. On the left side of the equal sign , write the price of his private island times the number of islands, 50 times 1, and add the price times the number of rocket crocs he plans to sell, 5 times 20.

On the right side, write the cost in gold bars, 10, times the unknown amount, x, which stands for the total number of diamonds he can put on his mom's ring.

Solving the Word Problem Example 1

Now, to solve the equation for x , you have to simplify the equation using PEMDAS and opposite operations .

Solving the Word Problem Example 2

But it's Mommy's 60th birthday and Dr. Evil decides he must buy her a ring with 60 diamonds. Wow! In addition to the island, how many rocket crocs will he need to sell? Let this now be x.

Now let's write another equation. Again, write the price of the island multiplied by the number of islands for sale and add the price of the rocket crocs times x on the left side of the equation.

Make this equal to the price of the gold bars times the number of diamonds Dr. Evil wishes to put on his mom's ring. Simplify the equation using PEMDAS and solve for the unknown variable using opposite operations .

Solving the Word Problem Example 3

What? You've gotta be kidding! He has only 55 of the rockets crocs! Oh boy, what can he do? Eureka! Dr. Evil realizes he can raise the price of the rocket crocs. Back to the equation...

This time, the unknown number, 'x' is the new price of the rocket crocs. So, again, write the price of the island multiplied by the number of islands for sale and add the price of the rocket crocs, our new x, times 55, the number of rocket crocs he has, on the left side of the equation.

Set this equal to the value and the number of diamonds on the ring. Simplify and solve.

Great, now that he's finished with his calculations, he can start the sale. Oh no! One of the robotic rocket crocs just exploded! Maybe Dr. Evil can buy smaller diamonds?

Solving Equations: Word Problems exercise

Examine the number of diamonds dr. evil can buy for his mom's ring..

The solution is a natural number.

For his mom's birthday present, Dr. Evil would like to buy a ring. His mom loves diamonds, but one diamond costs $10$ gold bars. Dr. Evil considers selling one island for $50$ gold bars and $20$ rocket crocodiles for $5$ gold bars each.

After he sells the island and rocket crocs, how many diamonds can Dr. Evil buy for his mom?

The number of diamonds is unknown. So we assign the variable $x$ to it. We know the cost of one diamond is $10$ gold bars. So $x$ diamonds cost $10\times x$ gold bars.

$50\times 1+5\times 20=10\times x$

Oh, but it's his mom's $60\text{th}$ birthday. $15$ diamonds aren't enough, he thinks. Now he has to look for some other possibilities...

Decide which equation best describes each of the given situations.

First, determine the information we already know, and then assign the variable $x$ to the unknown value.

For example, if one rocket crocodile costs $5$ gold bars $4$ rocket crocodiles cost $4\times 5=20$ gold bars.

You just have to multiply the number of rocket crocodiles by the price per crocodile.

Don't forget Dr. Evil sells his island for $50$ gold bars.

To solve word problems, you must recognize what values are given and what is unknown.

To write an equation, we assign the the variable $x$ to our unknown value.

Determine how many hours Dr. Evil's assistant can work.

Unfortunately, one of the rocket crocodiles exploded, and Dr. Evil's lab is completely destroyed.

Assistants are very expensive, $4$ gold coins an hour. The number of hours is unknown. So we assign the variable $x$ to this unknown number.

Set up equations and solve for the unknown values.

First, assign the variable $x$ to the unknown.

Formulate an equation with the things Lara sells on one side of the equal sign and the total money she needs for the teacups on the other side.

Use PEMDAS to simplify the equation.

To isolate the variable $x$, you have to perform opposite operations.

$\begin{array}{rclcl} 29\times x &=& 1.50\times 30+2.50\times 40 & &|\text{ PEMDAS}\\ 29x&=&45+100\\ 29x&=&~145 & &|\text{ Combine Like Terms}\\ \color{#669900}{\div29}&&\color{#669900}{\div29}& &| (\times~\leftrightarrow~\div) \text{ Opposite Operations}\\ x&=&~~~~~5 \end{array}$

$x=5$ means Lara can buy $5$ new teacups.

Second Example In the second example, the price of each book is unknown. We have $1.50\times 30+x\times 40$ on one side and the cost of 6 new teacups on the other side: $37.50\times 6$. Now we have enough information to solve the equation.

$\begin{array}{rclcl} 37.50\times 6&=&2\times 30+4\times x&&|\text{PEMDAS}\\ 225&=&45+40x\\ \color{#669900}{-45}&&\color{#669900}{-45}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ 180&=&~40x\\ \color{#669900}{\div40}&&\color{#669900}{\div40}&&| (\times~\leftrightarrow~\div) \text{ Opposite Operations}\\ x&=&~4.5 \end{array}$

If Lara sells each book for $\$4.50$, she can buy the $6$ new teacups she wants.

Third Example In our last example, the number of books is unknown. We have $1.50\times 30+4\times x$ on one side and the cost of 6 new teacups on the other: $37.5\times 6$. Let's solve it!

$\begin{array}{rclcl} 25\times 8&=&2\times 30+4\times x&&|\text{PEMDAS}\\ 200&=&60+4x\\ \color{#669900}{-60}&&\color{#669900}{-60}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ 140&=&~~~4x\\ \color{#669900}{\div4}&=&\color{#669900}{\div4}&&| (\times~\leftrightarrow~\div) \text{ Opposite Operations}\\ x&=&~~~35 \end{array}$

$x=35$ means Lara has to sell $35$ books.

Summarize what you know about opposite operations.

To solve an equation like $x+3=7$, you must isolate the variable:

$\begin{array}{rclcl} x+3&=&~~~7\\ \color{#669900}{-3}&&\color{#669900}{-3}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ x&=&~~~4. \end{array}$

Think of an equation like a scale in balance.

Removing weight on one side leads to an imbalance, so to stay balanced, you will need to remove it on the other side, too.

To solve an equation, the goal is to get a solution for the variable. That means that you should isolate the variable on one side of the equation.

We can use Opposite Operations to isolate the variable $x$.

Solve the following equations.

Some solutions do not equal $-1$, $1$ or $2$.

For every equation, we first combine like terms using PEMDAS. Then we isolate the variable $x$ using opposite operations.

Equation One

$\begin{array}{rclcl} 5x-3x+2&=&~~~4\\ 2x+2&=&~~~4&&|\text{ Combine Like Terms}\\ \color{#669900}{-2}&&\color{#669900}{-2}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ 2x&=&~~~2\\ \color{#669900}{\div2}&&\color{#669900}{\div2}&&| (\times~\leftrightarrow~\div) \text{ Opposite Operations}\\ x&=&~~~2 \end{array}$

Equation Two

$\begin{array}{rclcl} 2x+3\times(x-1)&=&~~~7&&|\text{ Distributive Property}\\ 2x+3x-3&&~~~7\\ 5x-3&=&~~~7&&|\text{ Combine Like Terms}\\ \color{#669900}{+3}&&\color{#669900}{+3}&&| (-~\leftrightarrow~+) \text{ Opposite Operations}\\ 5x&=&~10\\ \color{#669900}{\div5}&&\color{#669900}{\div5}&&| (\times~\leftrightarrow~\div) \text{ Opposite Operations}\\ x&=&~~~2 \end{array}$

Equation Three

$\begin{array}{rclcl} 5x+2&=&~~~4\times(x+1)\\ 5x+2&&4x+4&&|\text{ Distributive Property}\\ \color{#669900}{-2}&&\color{#669900}{-2}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ 5x&=&4x+2\\ \color{#669900}{-4x}&&\color{#669900}{-4x}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ x&=&~~~2 \end{array}$

Equation Four

$\begin{array}{rclcl} 2x+4&=&4\times(x-1)+10&&|\text{ Distributive Property}\\ 2x+4&=&4x-4+10\\ 2x+4&=&4x+6&&|\text{ Combine Like Terms}\\ \color{#669900}{-4}&&\color{#669900}{-4}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ 2x&=&4x+2\\ \color{#669900}{-4x}&&\color{#669900}{-4x}&&| (+~\leftrightarrow~-) \text{ Opposite Operations}\\ -2x&=&~~~2\\ \color{#669900}{\div(-2)}&&\color{#669900}{\div(-2)}&&| (\times~\leftrightarrow~\div) \text{ Opposite Operations}\\ x&=&-1 \end{array}$

The other two equations have solutions that do not equal $-1$, $1$ or $2$.

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  5. 3 Ways to Solve Word Problems Requiring Quadratic Equations

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  6. Solving Word Problems Involving Linear Equations Math Worksheets

    solving equations word problems

VIDEO

  1. 3.6: Equations and Problem Solving

  2. Solving Equations--Word Problems Notes Part #3

  3. Solving Simple Linear Equations| Worded Problems

  4. Points Activity Problem 3

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  6. Solving Equations Word Problems

COMMENTS

  1. Algebraic word problems

    Algebraic word problems are questions that require translating sentences to equations, then solving those equations. The equations we need to write will only involve basic arithmetic operations and a single variable. Usually, the variable represents an unknown quantity in a real-life scenario.

  2. Equation Word Problems Worksheets

    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. Two-Step Equation Word Problems: Fractions and Decimals

  3. Word Problems Calculator

    To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform.

  4. Two-step equations word problems (practice)

    Two-step equations word problems CCSS.Math: 7.EE.B.4, 7.EE.B.4a Google Classroom You might need: Calculator Mindy and Troy combined ate 9 9 pieces of the wedding cake. Mindy ate 3 3 pieces of cake and Troy had \dfrac14 41 of the total cake. Write an equation to determine how many pieces of cake (c) (c) there were in total.

  5. Linear equations word problems

    Linear equations word problems. CCSS.Math: HSF.LE.B.5. Google Classroom. Ever since Renata moved to her new home, she's been keeping track of the height of the tree outside her window. H H represents the height of the tree (in centimeters), t t years since Renata moved in. H=210+33t H = 210+ 33t.

  6. Browse Printable Solving Equation Worksheets

    Students will draw on their algebra knowledge to answer word problems by writing and then solving two-step equations. 7th grade. Math. Worksheet. Solve One-Step Addition and Subtraction Equations ... Write and Solve Equations From Word Problems. Help learners understand and solve real-world problems using algebraic reasoning with these mixed ...

  7. 1.20: Word Problems for Linear Equations

    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 We solve this for x : x = 60 5 = 12 b) If 5 is subtracted from twice an unknown number, the difference is 13. Find the number. Solution: Translating the problem into an algebraic equation gives:

  8. Algebra Word Problems

    Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem. The following key words will help you write equations for Algebra word problems: Addition altogether increase more plus sum total combine subtraction difference decrease less fewer reduce minus multiplication per times

  9. Algebra Topics: Introduction to Word Problems

    Here's how: First, replace the cost of a family pass with our variable f. f equals half of $8 plus $25. Next, take out the dollar signs and replace words like plus and equals with operators. f = half of 8 + 25. Finally, translate the rest of the problem. Half of can be written as 1/2 times, or 1/2 ⋅ : f = 1/2 ⋅ 8 + 25.

  10. Variables and equations word problem worksheets

    These word problems involve variables and equations. Students write out a full equation with the suggested variable representing the unknown, then solve the equation. Underlying operations include addition, subtraction, multiplication and division of whole numbers and fractions. Worksheet #1 Worksheet #2 Worksheet #3 Worksheet #4.

  11. Algebra

    Algebra - Solving Equations (word problem) yaymath 243K subscribers Subscribe 127K views 6 years ago Yay Math in Studio presents a single word problem, the likes of which students around...

  12. PDF SYSTEM OF EQUATIONS-WORD PROBLEMS #2 Directions system of equations

    Name_____ SYSTEM OF EQUATIONS-WORD PROBLEMS #2 . Directions: Find the answers to each situation below by setting up and solving a system of equations.. 1) Ben bought a total of 11 pizzas for his class party. The amount of large pizzas was one less than twice the amount of medium pizzas. How many large and how many medium pizzas did he buy?

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

    Find the key information in the word problem that can help you define the variables. Define two variables: x x and y 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. Systems of Equations Word Problems

  14. Systems of equations word problems

    Systems of equations word problems (with zero and infinite solutions) Systems of equations with elimination: TV & DVD Systems of equations with elimination: apples and oranges Systems of equations with substitution: coins Systems of equations with elimination: coffee and croissants Systems of equations: FAQ Math > Algebra 1 > Systems of equations >

  15. How to Solve Word Problems in Algebra

    Determine what you are asked to find. In many problems, what you are asked to find is presented in the last sentence. This is not always true, however, so you need to read the entire problem carefully. Write down what you need to find, or else underline it in the problem, so that you do not forget what your final answer means. In an algebra word problem, you will likely be asked to find a ...

  16. Quadratic Equations: Very Difficult Problems with Solutions

    Problem 6 sent by Κυριάκος There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original's minus one, and whose last two digits are the same and equal to the half of the original's.

  17. Two-step equation word problem: garden (video)

    Now we can use the inverse property of multiplication (which is division) to solve for W. We can divide 6 times W by 6; and we also have to do that to the other side so we have 6W/6 = 60/6; then we get the value of W which equals 10 ft. Now we can substitute what we know about length (L). We know that L is equal to 2W.

  18. Solving Word Problems: Steps & Examples

    Learn the steps used in solving word problems which include visualizing the problem, writing the equation, and solving the equation with examples of how it's done. Updated: 10/05/2021 Create an ...

  19. Rational equations word problem: combined rates (example 2)

    For the larger hose 1/55 of the pool will be filled in a minute. The smaller hose would then be 1/x per minute. Now we can add both hoses together (1/x + 1/55) and set it equal to how much of the pool is filled when both are working (1/35) 1/x +1/55 = 1/35 and solve. 1/x=1/35-1/55. 1/x=4/385.

  20. Writing Equations From Word Problems 6th Grade Worksheets Teaching

    Equations and Word Problems Task Cards: Use these task cards instead of worksheets to get students practicing solving equations (8 problems), writing equations from real world situations (12 problems), and generating their own story problems represented by a given equation (4 problems). These task cards come in both printable AND digital formats.

  21. Microsoft Math Solver

    Microsoft Math Solver - Math Problem Solver & Calculator Get step-by-step solutions to your math problems Try Math Solver Type a math problem Solve Quadratic equation x2 − 4x − 5 = 0 Trigonometry 4sinθ cosθ = 2sinθ Linear equation y = 3x + 4 Arithmetic 699 ∗533 Matrix [ 2 5 3 4][ 2 −1 0 1 3 5] Simultaneous equation { 8x + 2y = 46 7x + 3y = 47

  22. Equations for solving Word Problems

    Using Other Types of Equations to Solve Word Problems There are many types of equations besides the first - order equations with one variable. We will discuss a few examples where two other types of equations (second - order equations with one variable and first - order equations with two variables) are used in word problems.

  23. Solving Equations: Word Problems

    Solving the Word Problem Example 1 Now, to solve the equation for x, you have to simplify the equation using PEMDAS and opposite operations. Here, multiplication is first. 50 · 1 = 50 and 5 · 20 = 100. Now we have to add: 50 + 100 = 150. 150 = 10x. To isolate x, we use the opposite operation here. We divide both sides by 10. x = 15.