linear equations word problems pdf

Linear Equation Word Problems Worksheet

Students will practice solving linear equation word problems.

Example Questions

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This is a 4 part worksheet:

• Part I Model Problems
• Part II Practice
• Part III Challenge Problems
• Part IV Answer Key
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Linear Equations Word Problems Worksheet with Solutions

The first equation family that many students learn about is linear equations. Linear equations have many applications in the real-world, which can make for a really great set of word problems! 

As a student studying algebra, you will encounter many linear equations word problems. That’s why I have put together this linear equations word problems worksheet with solutions!

My hope is that this linear equation word problems worksheet and answer key help you deepen your understanding of linear equations and linear systems!

What Are Linear Equations?

A linear equation is an  algebraic equation  where the highest power on the variable is one. When graphed, a linear equation will produce a straight line.

There are a few ways that we can write linear equations, with two of the most common being slope-intercept form and standard form .

Slope-intercept form is best way to identify the slope of the line and the y-intercept of the line. In general, the equation of a line in slope-intercept form is written as:

In this form,  a  represents the slope of the line and  b  represents the y-intercept of the line.

Equations of lines in standard form are easy to recognize because it is a uniformly recognized form of a line. Standard form allows for easy comparison of coefficients. When two linear equations are in standard form, you can quickly compare the coefficients of x and y.

In general, the standard form of a line is written as:

$$Ax+By=C$$

Note that A and B do not represent the slope of the line or the y-intercept in this form. Instead, A and B are simply constants.

Solving Linear Equations

Any set of word problems relating to linear equations will ask you to solve an equation of some sort. However, there are many different types of solving equations problems that you will encounter as you explore linear equations word problems.

Let’s take a look at a few different types to make sure you know what to expect when you check out the linear equation word problems worksheet with solutions below.

Solving Two-Step Equations and Multi-Step Equations

One of the simplest equation problems that you can solve is a two-step equation. A two-step equation requires you to perform just two steps in order to determine the unique solution to the linear equation.

The first step is to identify the side of the equation with the unknown variable. Your goal will be to isolate this variable (or get it by itself). Consider the following example:

In this example, the variable is on the right-hand side of the equation. To isolate x, we “undo” the operations on the right-hand side of the equation using inverse operations. This just means doing whatever the opposite operation is.

We can apply  order of operations  in reverse to start with the subtraction and then deal with the multiplication. Adding 5 to both sides and then dividing by 2 will result in:

$$\begin{split} 9+5&=2x-5+5 \\ \\ 14 &= 2x \\ \\ \frac{14}{2} &= \frac{2x}{2} \\ \\ 7 &= x \end{split} $$

This shows that the value of the unknown variable here is x = 7. This is a unique solution that will satisfy the equation.

If you want to learn more about finding the solution of linear equations and explore multi-step equations that use the distributive property, check out  these equation solving worksheets !

Systems of Linear Equations

Systems of linear equations are another type of problem that you will see on the linear equations word problems worksheet linked below. A system of linear equations involves two (or more!) linear equations that intersect in some way. 

There are a few different ways that linear equations can intersect :

• Once  at a single point of intersection
• Never  as a result of the lines being parallel 
• Always  as a result of the lines being on top of one another

When solving a system of linear equations, your goal is to determine both unknown variables. If the lines intersect, the solution to the system will be the point of intersection for the lines.

When given a linear equation, we can find the point of intersection between it and a second equation using a few different methods. I made a  video on the substitution method  and a  video on the elimination method  to help you understand these strategies for solving systems before you apply them to word problems involving systems of equations.

What Are Linear Equation Word Problems?

A linear equation word problem involves a real-world situation or scenario that can be solved by setting up and solving linear equations. The equations that are used model the relationships between different quantities in the real-world scenario. 

The topics of these problems vary, ranging from applications in science and physics (ie. calculate the speed of the boat) to business applications (ie. how many sales are required to break even?). The problems that you encounter will also vary in depth and difficulty. 

In my teaching experience, students tend to struggle with word problems because it isn’t always immediately clear what is being asked.  I have seen many students feel very confident in their equation solving skills, yet they still struggle when it comes to solving linear equation word problems.

One reason for this is that you aren’t always given equations from the start while solving word problems.

Tips for Solving Linear Equation Word Problems

During my time as a high school math teacher, I have come across a few tips that I think will help you solve linear equation word problems successfully.

To begin, the first step should always be to define two variables. Read the question carefully and think about the quantities involved. Use variables to represent these quantities.

The second step should be writing an equation that models the scenario. Depending on the problem, you may need to write a second equation as well.

Lastly, think about what the problem is asking you to find. 

For example, are you looking for the values of two unknown variables? If so, you are likely going to be setting up and solving a system of linear equations.

​If you are being asked for the value of a single variable, the chances are you will be solving a single linear equation.

Now that you have a basic understanding of the concepts involved in solving linear equations word problems, it’s time to try a few!

My goal here is to provide you with a worksheet that you can use to practice and feel confident that you understand linear equations word problems!

While I was writing this worksheet, I made sure to include a wide variety of problems that range in difficulty. You will see a few simpler problems involving a two-step equation or multi-step equations, but you will also see a few problems that involve systems of linear equations.

After solving each word problem, be sure to check the answer key to verify that you fully understand the process used to set up the problem and solve it. Reflecting on your understanding is an important part of developing comfort with any given math concept!

Click below to download the linear equations word problems worksheet with solutions!

Using This Linear Equation Word Problems Worksheet

Being able to read a  real-world algebra problem  and set up a linear equation (or a system of linear equations) to solve it is a very challenging skill. In my experience as a math teacher, many students struggle with this concept, even if they fully understand the mathematics that the problem requires.

This is the main reason that I put together this linear equation word problems worksheet with solutions. My goal is to provide you with a set of word problems that you can use to check your understanding of solving linear equations in the real-world.

I hope you found this practice worksheet helpful as you continue your studies of algebra and linear equations!

If you are looking for more linear equations math worksheets in PDF formats, check out my collection of  solving linear inequalities worksheets  and this  linear inequality word problems worksheet .

Did you find this linear equation word problems worksheet with solutions helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!

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1.20: Word Problems for Linear Equations

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

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

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

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

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

\[7-3 x\nonumber\]

c) 8 less than a number.

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

\[x-8\nonumber\]

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

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

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

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

Simplifying this expression gives:

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

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

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

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

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

Example 18.2

Solve the following word problems:

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

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for \(x\) :

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

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

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

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

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

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

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

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

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

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

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

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

Solution: We have the equation:

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

Subtracting \(x\) gives

\[4x = 12.\nonumber\]

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

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

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

\[x + 9 = 3x − 7.\nonumber\]

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

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

Then we subtract \(x:\)

\[16 = 2x.\nonumber\]

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

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

Example 18.3

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

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

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

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

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

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

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

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

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

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

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

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

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

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

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

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

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

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

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

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

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

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

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

Distributing and collecting like terms give

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiplying both sides by 10 gives

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

Subtracting 64 from both sides gives

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

and dividing by 20 gives

Exit Problem

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

Linear Equations Word Problems Worksheets

Linear equations are equations that have two variables and are a straight line when graphed, based on their slope and y-intercept . Hence,linear equations word problems worksheets have a variety of word problems that help students practice key concepts and build a rock-solid foundation of the concepts.

Benefits of Linear Equations Word Problems Worksheets

Linear equations word problems worksheets are a great resource for students to practice a large variety of word type questions. These worksheets are supported by visuals which help students get a crystal clear understanding of the linear equations word type topic. The variety of problems that these worksheets offer helps students approach these concepts in an engaging and fun manner.

Linear equations word problems worksheets come with visual simulation for students to see the problems in action, an answer key that provides a detailed step-by-step solution for students to understand the process better, and a worksheet with detailed solutions.

Download Linear Equations Word Problems Worksheet PDFs

These math worksheets should be practiced regularly and are free to download in PDF formats.

☛ Check Grade wise Linear Equations Word Problems Worksheets

• 8th Grade Linear Equations Worksheets