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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?

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.

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 .

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by cpfaffinator

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

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

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

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

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

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

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

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

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

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

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

What does the y variable represent in our equation?

Years since the trees were planted

Number of trees planted

Radius of trees' roots

Height of the trees

Related Lessons

Translate to a linear equation.

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

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

• Find the y variable, or output. What is the thing you're trying to find? This will often be a price, or an amount of time, or something else countable that depends on other things.
• Find the x variable, or input. What is affecting the price, or amount of time, etc.?
• Find the slope. What's the rate in the problem?
• Find the y-intercept. Is there anything that's added or subtracted on top of the rate, no matter what what x is?
• Plug all the numbers you know into y=mx+b !

Wally and Cobb are starting a catering business. They rent a kitchen for $350 a month, and charge $75 for each event they cater. If they cater 12 events in a month, how much do they profit?

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

The name of the business

Word Problems with Linear Relationships

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

Image source: by Anusha Rahman

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