linear equations word problems and answers

Word Problems Linear Equations

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$\textbf{1)}$ 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.

one linear equation being substituted into a second equation

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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Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.

Solution: Then the other number = x + 9 Let the number be x. Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9) ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12 Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72 ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.

Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts. Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40 The two parts are 15 and 25.

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.

7. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair. Solution: The table cost $ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165.

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

● Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

● Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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Linear Equation Word Problems Worksheet

Students will practice solving linear equation word problems.

$Download this web page as a pdf with answer key$

Example Questions

linear equations word problems and answers

Other Details

This is a 4 part worksheet:

Part I Model Problems
Part II Practice
Part III Challenge Problems
Part IV Answer Key
Interactive Slope : explore the formula of slope by clicking and dragging two points.

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

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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

Linear Equation Word Problems - Examples & Practice - Expii

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

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!

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SAT Mathematics : Solving Linear Equations in Word Problems

Study concepts, example questions & explanations for sat mathematics, all sat mathematics resources, example questions, example question #31 : solving word problems.

Jackie plans to buy one video game and a number of hardcover books. The video game costs $40 and each book costs $30. If she must spend at least $120 in order to get free shipping, what is the minimum number of books she must buy in order to get free shipping?

Since you know that Jackie will purchase exactly one video game for $40, you can set up your equation (or inequality) here as:

$linear equations word problems and answers$

Now you can subtract 40 from both sides:

$linear equations word problems and answers$

And when you divide both sides by 30 you'll see that you have a number between 2 and 3. Note that you do not have to do the full decimal calculation, because you cannot buy part of a book! So since she can't buy 2.67 books, the minimum number she can purchase is 3.

Example Question #32 : Solving Word Problems

On the first day of the week, a bakery had an inventory of 450 loaves of bread. It bakes 210 loaves of bread and sells 240 loaves of bread each day that it is open, and then closes for a baking day when it runs out of loaves. How many days can it be open before it must close for a baking day?

If the bakery bakes 210 loaves of bread and sells 240 loaves of bread, then that means that, total, it loses 30 loaves of bread per day. Since you know that it starts with 450 loaves of bread, you can use this information to write a linear equation relating the number of loaves of bread left with how many days it has been since the bakery has closed for a baking day. It should look like:

B = 450 - 30d

Where B represents the number of loaves left and d represents the number of days since the bakery’s last baking day.

In order for the bakery to need to close for a baking day, the number of loaves left must equal 0. If you substitute in B = 0, you get: 0 = 450−30d.

Now you can solve: add 30d to both sides to get:

And then divide both sides by 30 to get d = 15

Example Question #1 : Solving Linear Equations In Word Problems

Katharine currently has $10,000 saved for a down payment on purchasing her first house, which she can do when her savings has reached $50,000. Each month she earns $6,500 but incurs $4,000 in expenses. If her earnings and expenses remain constant, how many months will it take until she has reached her savings goal?

To turn this problem into an equation, you can start by putting Katharine's goal of $50,000 on one side of the equation, and then arranging all of the inputs to that goal (ways she earns money) and impediments (ways she loses money) on the other. If you use m to represent the number of months, an initial equation would look like:

$50,000 = $10,000 + $6,500m - $4,000m

Which, qualitatively, is that her goal of $50,000 is equal to the $10,000 she has plus the $6,500 she earns each month, minus the $4,000 she loses each month.

Now you can combine like terms to simplify the equation:

$40,000 = $2,500m

And then divide 40000 by 2500 (which simplifies to 400 divided by 25) to solve:

m = 16 months

Example Question #34 : Solving Word Problems

$linear equations word problems and answers$

Then you'll need to find a common denominator of 6, and rewrite what you have to reflect that common denominator:

$linear equations word problems and answers$

You can then combine like terms on the right-hand side of the equation:

$linear equations word problems and answers$

And then multiply both sides by 6 to finish the problem:

$linear equations word problems and answers$

Example Question #35 : Solving Word Problems

A wood beam 48 inches in length is cut into two smaller beams. If one of the new beams is 20 inches longer than the other, what is the length of the longer beam?

$linear equations word problems and answers$

Example Question #36 : Solving Word Problems

$linear equations word problems and answers$

While it is tempting to set up and solve an equation for this problem, the algebra for this problem may be difficult to translate quickly and accurately. Because you are dealing with relatively simple numbers that add up to a total, you should see that this should be an easy problem to backsolve.

$linear equations word problems and answers$

Alternatively, you could set up the algebra:

$linear equations word problems and answers$

Example Question #37 : Solving Word Problems

Eight times one-third of a number is one greater than five times one half of that number. What is that number?

$linear equations word problems and answers$

This problem can be solved algebraically but is likely solved more quickly by simply back solving. Your algebra would set up as:

$linear equations word problems and answers$

So start with 6. Eight times one-third of 6 is 8(2) = 16. Is that one greater than five times one half of 6? It is. Five times one half of 6 is 5(3) = 15. The relationship works, so 6 is correct.

Example Question #38 : Solving Word Problems

$linear equations word problems and answers$

Importantly, note that if 60 hybrids are sold, that means that 60 of the total of 480 cars are sold. That affects both the number of hybrids AND the number of total cars, which is now reduced to 420. So your new proportion of hybrids is the 60 hybrids out of the 420 total cars:

$linear equations word problems and answers$

Example Question #39 : Solving Word Problems

Five years ago, Juliet was three times as old as Will. If Juliet is currently twice as old as Will, how old is Juliet?

$linear equations word problems and answers$

Example Question #40 : Solving Word Problems

On the same day, Devin bought a car for $40,000 that would then decrease in value by $1200 per year, and Anita purchased a piece of real estate for $6,000 that would then increase in value by $500 per year. After how many years will Anita’s land be worth as much as Devin’s car?

$linear equations word problems and answers$

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Word Problems on Linear Equations | How to Solve Linear Equations Word Problems?

We have provided several problems that involve relations among known and unknown numbers and can be put in the form of linear equations. Those equations can be stated in words and it is the main reason we prefer these Word Problems on Linear Equations. You can practice as many types of questions as you want to get an expert in this concept. For better understanding, we even listed linear equations examples with solutions.

Steps to Solve Word Problems on Linear Equations

Below are the simple steps to solve the linear equations word problems. Follow these instructions and solve the questions carefully.

Read the problem carefully and make a note of what is given in the question and what is required.
Denote the unknown things as the variables like x, y, z, a, b, . . .
Translate the given word problem into mathematical statements.
Form the linear equations in one variable by using the conditions provided in the question.
Solve the unknown parameters from the equation.
Verify the condition with the obtained answer ad cross check whether it is correct or not.

Linear Equations Examples with Answers

A motorboat goes downstream in the river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water?

Let the speed of the boat in still water = x km/hr

Speed of the boat down stream = (x + 3) km/hr

Time taken to cover the distance = 5 hrs

Therefore, distance covered in 5 hrs = (x + 3) x 5

Speed of the boat upstream = (x – 3) km/hr

Time taken to cover the distance = 6 hrs

Therefore, distance covered in 6 hrs – (x – 3) x 6

Therefore, the distance between the two coastal towns is fixed, i.e., the same.

As per the question

5(x + 3) = 6(x – 3)

5x + 15 = 6x – 18

15 + 18 = 6x – 5x

Required speed of the boat is 33 km/hr

The perimeter of a rectangular swimming pool is 144 m. Its length is 2 m more than twice its width. What are the length and width of the pool?

Let l be the length of the swimming pool, w be the width of the swimming pool.

According to the question,

length l = 2w + 2

The perimeter of swimming pool = 144 m

2l + 2w = 144

Substitute l = 2w + 2

2(2w + 2) + 2w = 144

4w + 4 + 2w = 144

6w = 144 – 4

w = 140 / 6

Then, the length is

l = 2(23.3) + 2

Hence, the length and width of the rectangular swimming pool is 48.6 m, 23.3 m.

The sum of three consecutive even numbers is 126. What are the numbers?

Let the first even number be x, the second number be (x + 2), the third number be (x + 4).

According to the question, the sum of consecutive even numbers is 126.

First Number + Second Number + Third Number = 126

x + (x + 2) + (x + 4) = 126

3x + 6 = 126

Subtract 6 from both sides.

3x + 6 – 6 = 126 – 6

Divide both sides by 3.

3x / 3 = 120 / 3

The first number is 40, the second number is (x + 2) = 40 + 2 = 42, third number is (x + 4) = 40 + 4 = 44.

Hence, the three consecutive even numbers are 40, 42, 44.

When five is added to three more than a certain number, the result is 19. What is the number?

Let us take the number as x.

Add 5 to the three more than a certain number.

5 + x + 3 = 19

Subtract 8 from both sides of the equation.

x + 8 – 8 = 19 – 8

So, the number is 11.

Eleven less than seven times a number is five more than six times the number. Find the number?

Let the number be x.

11 less than the seven times a number is five more than six times the number.

7x – 11 = 6x + 5

7x – 6x = 5 + 11

Hence the required number is 16.

Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the first angle. Find the measure of the angles.

Let the triangle be ∆ABC.

So, ∠A = ∠B and ∠C = ∠A – 12 degrees

The sum of three angles of a triangle = 180 degrees

∠A + ∠B + ∠C = 180

∠A + ∠A + ∠A – 12 = 180

3∠A – 12 = 180

Add 12 to both sides of the equation.

3∠A – 12 + 12 = 180 + 12

3∠A / 3 = 192/3

Hence, the first and second angles of the triangle are 64 degrees, 64 degrees and the third angle is 64 – 12 = 52 degrees.

The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions.

Let the rectangle width is w.

Length of rectangle l = w + 15 cm

Given that, the perimeter of a rectangle is 150 cm

2l + 2w = 150

Substitute l = w + 15 cm in above equation.

2(w + 15) + 2w = 150

2w + 30 + 2w = 150

4w + 30 = 150

Subtract 30 from both sides.

4w + 30 – 30 = 150 – 30

Divide both sides by 4.

4w/4 = 120/4

Hence, the rectangle width is 30 cm, the length is (30 + 15) = 45 cm.

If Mr. David and his son together had 220 dollars, and Mr. David had 10 times as much as his son, how much money had each?

Let Mr. David’s son has x dollars.

The amount at Mr. David = 10x dollars

Given that, Mr. David and his son together had 220 dollars

x + 10x = 220

Divide both sides by 11.

x = 220 / 11

Hence, Mr. David has 20 x 10 = 200 dollars and his son has 20 dollars.

Linear systems – word problems exams for teachers

Linear systems – word problems worksheets for studets.

Basic Mathematical Operations

Addition and subtraction
Multiplying
Divisibility and factors
Order of operations
Greatest common factor
Least common multiple
Squares and square roots
Naming decimal places
Write numbers with words
Verbal expressions in algebra
Rounding numbers
Convert percents, decimals and fractions
Simplifying numerical fractions
Proportions and similarity
Calculating percents
The Pythagorean theorem
Quadrilaterals
Solid figures
One-step equations
One-step equations – word problems
Two-step equations
Two-step equations – word problems
Multi-step equations
Multiplying polynomials
Inequalities
One-step inequalities
Two-step inequalities
Multi-step inequalities
Coordinates
Graphing linear equations
The distance formula