word problems on linear equations in 2 variables

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.

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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Word Problems: Linear Equations in Two Variables

1. The perimeter of a rectangular garden is 20 m. If the length is 4 m more than the breadth, find the length and breadth of the garden.

Solution: Let the length of the garden be x m. Therefore, breadth of garden = (x - 4) m.

Since, perimeter is 20 m, so

2[x + (x - 4)] = 20

2(2x - 4) = 20

2x - 4 = 10

2x = 10 + 4 = 14

Hence, length = 7 m and breadth = 7 - 4 = 3 m.

Alternatively, you can solve the problem using two variables.

Let the length of garden = x m

Width of garden = y m

Therefore, x = y + 4 ... (1)

Also, perimeter is 20 m, therefore

2(x + y) = 20

x + y = 10 ... (2)

Solving (1) and (2), we get x = 7, y = 3

Hence, length = 7 m and breadth = 3 m

2. Asha is five years older than Robert. Five years ago, Asha was twice as old as Robert was then. Find their present ages.

Solution: Let present age of Asha be x years. Present age of Robert be y years.

Therefore, 

x - y = 5 ... (1)

5 years ago, Asha was (x - 5) years and Robert was (y - 5) years old. Therefore,

x - 5 = 2(y - 5)

x - 2y = - 5 ... (2)

Solving (1) and (2), we get y = 10 and x = 15

Hence, present age of Asha = 15 years and present age of Robert = 10 years.

3. Two places A and B are 100 km apart. One car starts from A and another from B at the same time. If they travel in the same direction, they meet after 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars. Assume that the speed of car at A is more than the speed of car at B.

Solution: Let speed of the car starting from A be x km/h and speed of the car starting from B be y km/h.

Therefore, the distance travelled by car at A in 5 hours = 5x km and the distance travelled by car at B in 5 hours = 5y km.

Since they meet after 5 hours when they travel in the same direction, the car at A has travelled 100 km more than the car at B. Therefore,

5x - 5y = 100

x - y = 20 ... (1)

When they travel towards each other, they meet after 1 hour. It means, total distance travelled by car at A and car at B in 1 hour is 100 km. Therefore,

x + y = 100 ... (2)

Solving (1) and (2), we get x = 60 and y = 40

Therefore, the speed of car at A = 60 km/h and the speed of car at B = 40 km/h.

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.

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How to Solve Two-variable Linear Equations Word Problems

Two-variable Linear Equations Word Problems refer to a specific category of mathematical problems that present a real-life scenario requiring the formulation and subsequent solution of two linear equations with two unknown variables. These problems demand a blend of comprehension skills and algebraic manipulation to arrive at a solution.

Step-by-step Guide to Solve Two-variable Linear Equations Word Problems

Here is a step-by-step guide to solving two-variable linear equations word problems:

Step 1: Initial Gathering of Thoughts:

Before anything else, immerse yourself in the story the problem tells. Without rushing to solve, familiarize yourself with the narrative.

Step 2: Key Element Identification:

Discover and underline essential entities (like quantities or amounts) and the relationships between them. These details will guide the formation of your equations.

Step 3: Symbolization of Unknowns:

Give a face (or a symbol) to the unknowns. Let’s take, for instance, \(x\) the first unknown and \(y\) for the second.

Step 4: Sculpting the Equations:

Use the relationships you’ve identified to craft the equations. Remember, the art lies in ensuring that the equations genuinely capture the essence of the story.

Step 5: Assembly of the System:

Place the two equations side-by-side, creating a picturesque system. This will help you visualize the interconnections.

Step 6: Selection of the Tackling Technique:

There are multiple ways to unmask the unknowns:

• Substitution Method: Isolate one of the variables and replace it in the other equation.
• Elimination Method: Adjust the coefficients such that adding or subtracting the equations eliminates one of the variables.
• Graphical Method: If you’re visual, sketch both equations on a graph. The point of intersection is your solution.

Step 7: Decoding the Equations:

Solve the jigsaw! Dive deep into the equations to discover the values of \(x\) and \(y\). Remember, accuracy is paramount.

Step 8: Intersecting Solutions with the Story:

Revisit the original narrative. Integrate your solutions (the values of \(x\) and \(y\)) back into the story to ensure everything aligns harmoniously.

Step 9: Reality Check:

Dive back in! Does your solution make sense in the context of the word problem? If your answers seem absurd or out of place, retrace your steps.

Step 10: Elation & Expression:

Once you’re confident, articulate your solution clearly, with emphasis on each step’s logic. Your journey through the problem is as essential as the solution itself.

Remember, in the cosmos of mathematics, every word problem is a tale waiting for its plot to be unraveled. Your task is not just to solve but to be the storyteller, unveiling each twist with precision and care. Happy solving!

by: Effortless Math Team about 9 months ago (category: Articles )

Effortless Math Team

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Mastering Grade 6 Math Word Problems The Ultimate Guide to Tackling 6th Grade Math Word Problems

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3.3: Applications of Linear Systems with Two Variables

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Learning Objectives

• Set up and solve applications involving relationships between two variables.
• Set up and solve mixture problems.
• Set up and solve uniform motion problems (distance problems).

Problems Involving Relationships between Two Variables

If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.

Example \(\PageIndex{1}\):

The sum of \(4\) times a larger integer and \(5\) times a smaller integer is \(7\). When twice the smaller integer is subtracted from \(3\) times the larger, the result is \(11\). Find the integers.

Begin by assigning variables to the larger and smaller integer.

Let \(x\) represent the larger integer.

Let \(y\) represent the smaller integer.

When using two variables, we need to set up two equations. The first sentence describes a sum and the second sentence describes a difference.

This leads to the following system:

\(\left\{ \begin{array} { l } { 4 x + 5 y = 7 } \\ { 3 x - 2 y = 11 } \end{array} \right.\)

Solve using the elimination method. To eliminate the variable \(y\) multiply the first equation by \(2\) and the second by \(5\).

\(\left\{ \begin{array} { l l } { 4 x + 5y = 7 } & { \stackrel { \times2 } { \Rightarrow } } \\ { 3 x -2y = 11 } & { \stackrel { \Rightarrow } { \times 5 } } \end{array} \right. \left\{ \begin{array} { l } { 8 x + 10 y = 14 } \\ { 15 x -10y = 55 } \end{array} \right.\)

Add the equations in the equivalent system and solve for \(x\).

\(\begin{aligned} 8 x \color{red}{+ 10 y} &\color{black}{=} 14 \\ \pm 15 x \color{red}{- 10 y} & \color{black}{=} 55 \\ \hline\\ 23x & = 99\\ x & = \frac{69}{23}\\x&=3 \end{aligned}\)

Back substitute to find \(y\).

\(\begin{aligned} 4 x + 5 y & = 7 \\ 4 ( \color{OliveGreen}{3} \color{black}{)} + 5 y & = 7 \\ 12 + 5 y & = 7 \\ 5 y & = - 5 \\ y & = - 1 \end{aligned}\)

The largest integer is \(3\) and the smaller integer is \(-1\).

Exercise \(\PageIndex{1}\)

An integer is \(1\) less than twice that of another. If their sum is \(20\), find the integers.

The two integers are \(7\) and \(13\).

www.youtube.com/v/LnzO1_J4X20

Next consider applications involving simple interest and money.

A total of \($12,800\) was invested in two accounts. Part was invested in a CD at a \(3 \frac{1}{8}\)% annual interest rate and part was invested in a money market fund at a \(4 \frac{3}{4}\)% annual interest rate. If the total simple interest for one year was \($465\), then how much was invested in each account?

Begin by identifying two variables.

Let \(x\) represent the amount invested at \(3 \frac{1}{8}\)% \(= 3.125\) % \(= 0.03125\).

Let \(y\) represent the amount invested at \(4 \frac{3}{4}\)% \(= 4.75\)% = \(0.0475\).

The total amount in both accounts can be expressed as

\(x+y=12,800\)

To set up a second equation, use the fact that the total interest was \($465\). Recall that the interest for one year is the interest rate times the principal \((I = prt = pr ⋅ 1 = p)\). Use this to add the interest in both accounts. Be sure r to use the decimal equivalents for the interest rates given as percentages.

\(\begin{aligned} \color{Cerulean} { interest\: from\: the\: C D\: +\: interest\: from\: the\: fund\: =\: total\: interest } \\ 0.03125 x \quad\quad\quad +\quad\quad\:\: \quad 0.0475 y \quad\quad\quad\quad = 465\quad\quad\quad\quad\:\: \end{aligned}\)

These two equations together form the following linear system:

\(\left\{ \begin{array} { c } { x + y = 12,800 } \\ { 0.03125 x + 0.0475 y = 465 } \end{array} \right.\)

Eliminate \(x\) by multiplying the first equation by \(-0.03125\).

Next, add the resulting equations.

\(\begin{aligned} \color{red}{- 0.03125 x}\color{black}{ -} 0.03125 y &= - 400 \\ \pm\:\: \color{red}{0.03125 x}\color{black}{ +} 0.0475 y &= 465 \\ \hline \\0.01625y &=65 \\ y & = \frac{65}{0.01635} \\ y & = 4,000 \end{aligned}\)

Back substitute to find \(x\).

\(\begin{aligned} x + y & = 12,800 \\ x + 4000 & = 12,800 \\ x & = 8,800 \end{aligned}\)

\($4,000\) was invested at \(4 \frac{3}{4}\)% and \($8,800\) was invested at \(3 \frac{1}{8}\)%.

Example \(\PageIndex{3}\):

A jar consisting of only nickels and dimes contains \(58\) coins. If the total value is \($4.20\), how many of each coin is in the jar?

Let \(n\) represent the number of nickels in the jar.

Let \(d\) represent the number of dimes in the jar.

The total number of coins in the jar can be expressed using the following equation:

Next, use the value of each coin to determine the total value \($4.20\).

\(\begin{aligned} \color{Cerulean} {value\: of\: nickels\: + \: value\: of\: dimes\: =\: total\:value} \\ 0.05 n\quad\quad\: +\quad\:\:\: 0.10 d\quad\quad\quad = 4.20\quad\quad\quad \end{aligned}\)

This leads us to the following linear system:

\(\left\{ \begin{array} { l } { n + d = 58 } \\ { 0.05 n + 0.10 d = 4.20 } \end{array} \right.\)

Here we will solve using the substitution method. In the first equation, we can solve for \(n\).

Substitute \(n = 58 − d\) into the second equation and solve for \(d\).

\(\begin{aligned} 0.05 ( \color{Cerulean}{58 - d}\color{black}{ )} + 0.10 d & = 4.20 \\ 2.9 - 0.05 d + 0.10 d & = 4.20 \\ 2.9 + 0.05 d & = 4.20 \\ 0.05 d & = 1.3 \\ d & = 26 \end{aligned}\)

Now back substitute to find the number of nickels.

\(\begin{aligned} n & = 58 - d \\ & = 58 - 26 \\ & = 32 \end{aligned}\)

There are \(32\) nickels and \(26\) dimes in the jar.

Exercise \(\PageIndex{2}\)

Joey has a jar full of \(40\) coins consisting of only quarters and nickels. If the total value is \($5.00\), how many of each coin does Joey have?

Joey has \(15\) quarters and \(25\) nickels.

www.youtube.com/v/41bxt_tThkA

Mixture Problems

Mixture problems often include a percentage and some total amount. It is important to make a distinction between these two types of quantities. For example, if a problem states that a \(20\)-ounce container is filled with a \(2\)% saline (salt) solution, then this means that the container is filled with a mixture of salt and water as follows:

In other words, we multiply the percentage times the total to get the amount of each part of the mixture.

Example \(\PageIndex{4}\):

A \(1.8\)% saline solution is to be combined and mixed with a \(3.2\)% saline solution to produce \(35\) ounces of a \(2.2\)% saline solution. How much of each is needed?

Let \(x\) represent the amount of \(1.8\)% saline solution needed.

Let \(y\) represent the amount of \(3.2\)% saline solution needed.

The total amount of saline solution needed is \(35\) ounces. This leads to one equation,

The second equation adds up the amount of salt in the correct percentages. The amount of salt is obtained by multiplying the percentage times the amount, where the variables \(x\) and \(y\) represent the amounts of the solutions. The amount of salt in the end solution is \(2.2\)% of the \(35\) ounces, or \(.022(35)\).

\(\begin{aligned} \color {Cerulean} { salt\: in\: 1.8} \%\: \color{Cerulean}{solution } + \color{Cerulean} { salt\: in \:} 3.2 \% \color{Cerulean} { \:solution } = \color{Cerulean} { salt\: in\: the\: end\: solution } \\ 0.018 x \quad\quad\quad+ \quad\:\quad 0.032 y\quad\quad\quad\quad = \quad\quad\quad0.022 ( 35 )\quad\quad\quad \end{aligned}\)

The algebraic setup consists of both equations presented as a system:

\(\left\{ \begin{array} { c } { x + y = 35 } \\ { 0.018 x + 0.032 y = 0.022 ( 35 ) } \end{array} \right.\)

Add the resulting equations together

\(\begin{aligned} - 0.018 x - 0.018 y &= - 0.63 \\ \pm\:\: 0.018 x + 0.032 y &= 0.77 \\ \hline \\0.014y &=0.14\\y&=\frac{0.14}{0.014}\\y&=10 \end{aligned}\)

\(\begin{aligned} x + y & = 35 \\ x + \color{OliveGreen}{10} & \color{Black}{=} 35 \\ x & = 25 \end{aligned}\)

We need \(25\) ounces of the \(1.8\)% saline solution and \(10\) ounces of the \(3.2\)% saline solution.

Example \(\PageIndex{5}\):

An \(80\)% antifreeze concentrate is to be mixed with water to produce a \(48\)-liter mixture containing \(25\)% antifreeze. How much water and antifreeze concentrate is needed?

Let \(x\) represent the amount of \(80\)% antifreeze concentrate needed.

Let \(y\) represent the amount of water needed.

The total amount of the mixture must be \(48\) liters.

The second equation adds up the amount of antifreeze from each solution in the correct percentages. The amount of antifreeze in the end result is \(25\)% of \(48\) liters, or \(0.25(48)\).

\(\begin{aligned} \color {Cerulean} { antifreeze\: in\: 80} \%\: \color{Cerulean}{concentrate } + \color{Cerulean} { antrifreeze\: in \: water} = \color{Cerulean} { antifreeze\: in\: the\: end\: mixture } \\ 0.018 x \quad\quad \quad\quad\quad+ \: \quad\:\quad 0.032 y\quad\quad\quad\quad = \quad\quad\quad\quad0.022 ( 35 )\quad\quad\quad\quad\quad \end{aligned}\)

Now we can form a system of two linear equations and two variables as follows:

\(\left\{ \begin{array} { c } { x + y = 48 } \\ { 0.80 x = 0.25 ( 48 ) } \end{array} \right. \Rightarrow \left\{ \begin{array} { l } { x + y = 48 } \\ { 0.80 x = 12 } \end{array} \right.\)

Use the second equation to find \(x\):

\(\begin{aligned} 0.80 x & = 12 \\ x & = \frac { 12 } { 0.80 } \\ x & = 15 \end{aligned}\)

\(\begin{aligned} x + y & = 48 \\ \color{OliveGreen}{15} + y & = 48 \\ y & = 33 \end{aligned}\)

We need to mix \(33\) liters of water with \(15\) liters of antifreeze concentrate.

Exercise \(\PageIndex{3}\)

A chemist wishes to create \(100\) ml of a solution with \(12\)% acid content. He uses two types of stock solutions, one with \(30\)% acid content and another with \(10\)% acid content. How much of each does he need?

The chemist will need to mix \(10\) ml of the \(30\)% acid solution with \(90\) ml of the \(10\)% acid solution.

www.youtube.com/v/NXbyJNE9mWw

Uniform Motion Problems (Distance Problems)

Recall that the distance traveled is equal to the average rate times the time traveled at that rate, \(D = r ⋅ t\). These uniform motion problems usually have a lot of data, so it helps to first organize that data in a chart and then set up a linear system. In this section, you are encouraged to use two variables.

Example \(\PageIndex{6}\):

An executive traveled a total of \(4\) hours and \(875\) miles by car and by plane. Driving to the airport by car, she averaged \(50\) miles per hour. In the air, the plane averaged \(320\) miles per hour. How long did it take her to drive to the airport?

We are asked to find the time it takes her to drive to the airport; this indicates that time is the unknown quantity.

Let \(x\) represent the time it took to drive to the airport. Let \(y\) represent the time spent in the air.

Fill in the chart with the given information.

Use the formula \(D = r \cdot t\) to fill in the unknown distances.

\(\begin{array} { l } { \text { Distance traveled in the car: } D = r \cdot t = 50 \cdot x } \\ { \text { Distance traveled in the air: } D = r \cdot t = 320 \cdot y } \end{array}\)

The distance column and the time column of the chart help us to set up the following linear system.

\(\left\{ \begin{array} { c } { x + y = \:4 \:\:\color{Cerulean}{\leftarrow total \:time\:traveled }} \\ { 50 x + 320 y = 875 \color{Cerulean}{\leftarrow total\: distance\: traveled } } \end{array} \right.\)

\(\begin{aligned} \color{red}{- 50 x}\color{black}{ -} 50 y& = - 200 \\ \pm\:\:\color{red}{ 50 x}\color{black}{ +} 320 y& = 875 \\ \hline\\270y&=675\\y&=\frac{675}{270}\\y&=\frac{5}{2} \end{aligned}\)

Now back substitute to find the time \(x\) it took to drive to the airport:

\(\begin{aligned} x + y & = 4 \\ x + \color{OliveGreen}{\frac { 5 } { 2 }} & \color{Black}{=} 4 \\ x & = \frac { 8 } { 2 } - \frac { 5 } { 2 } \\ x & = \frac { 3 } { 2 } \end{aligned}\)

It took her \(1 \frac{1}{2}\) hours to drive to the airport.

It is not always the case that time is the unknown quantity. Read the problem carefully and identify what you are asked to find; this defines your variables.

Example \(\PageIndex{7}\):

Flying with the wind, a light aircraft traveled \(240\) miles in \(2\) hours. The aircraft then turned against the wind and traveled another \(135\) miles in \(1 \frac{1}{2}\) hours. Find the speed of the airplane and the speed of the wind.

Begin by identifying variables.

Let \(x\) represent the speed of the airplane.

Let \(w\) represent the speed of the wind.

Use the following chart to organize the data:

With the wind, the airplane’s total speed is \(x + w\). Flying against the wind, the total speed is \(x − w\).

Use the rows of the chart along with the formula \(D = r ⋅ t\) to construct a linear system that models this problem. Take care to group the quantities that represent the rate in parentheses.

\(\left\{ \begin{array} { l } { 240 = ( x + w ) \cdot 2 \color{Cerulean}{\leftarrow distance\: traveled\: with \:the \:wind } } \\ { 135 = ( x - w ) \cdot 1.5 \color{Cerulean}{ \leftarrow distance\: traveled\: against\: the\: wind } } \end{array} \right.\)

If we divide both sides of the first equation by \(2\) and both sides of the second equation by \(1.5\), then we obtain the following equivalent system:

\(\left\{ \begin{array} { l } { 240 = ( x + w ) \cdot 2 } \quad\quad\overset{\div 2}{\Longrightarrow} \\ { 135 = ( x - w ) \cdot 1.5 \:\quad\underset{\div 1.5}{\Longrightarrow} } \end{array} \right. \quad \left\{ \begin{array} { l } { 120 = x + w } \\ { 90 = x - w } \end{array} \right.\)

Here \(w\) is lined up to eliminate.

\(\begin{aligned} x \color{red}{+ w}&\color{black}{ =} 120 \\ \pm x \color{red}{- w}&\color{black}{=} 90 \\ \hline\\2x &=210\\x & = \frac{210}{2} \\x& = 105\end{aligned}\)

Back substitute

\(\begin{aligned} x + w & = 120 \\ \color{OliveGreen}{105}\color{black}{ +} w & = 120 \\ w & = 15 \end{aligned}\)

The speed of the airplane is \(105\) miles per hour and the speed of the wind is \(15\) miles per hour.

Exercise \(\PageIndex{4}\)

A boat traveled \(27\) miles downstream in \(2\) hours. On the return trip, which was against the current, the boat was only able to travel \(21\) miles in \(2\) hours. What were the speeds of the boat and of the current?

The speed of the boat was \(12\) miles per hour and the speed of the current was \(1.5\) miles per hour.

www.youtube.com/v/EvdJQTFSUSs

Key Takeaways

• Use two variables as a means to simplify the algebraic setup of applications where the relationship between unknowns is unclear.
• Carefully read the problem several times. If two variables are used, then remember that you need to set up two linear equations in order to solve the problem.
• Be sure to answer the question in sentence form and include the correct units for the answer.

Exercise \(\PageIndex{5}\)

Set up a linear system and solve.

• The sum of two integers is \(45\). The larger integer is \(3\) less than twice the smaller. Find the two integers.
• The sum of two integers is \(126\). The larger is \(18\) less than \(5\) times the smaller. Find the two integers.
• The sum of two integers is \(41\). When \(3\) times the smaller is subtracted from the larger the result is \(17\). Find the two integers.
• The sum of two integers is \(46\). When the larger is subtracted from twice the smaller the result is \(2\). Find the two integers.
• The difference of two integers is \(11\). When twice the larger is subtracted from \(3\) times the smaller, the result is \(3\). Find the integers.
• The difference of two integers is \(6\). The sum of twice the smaller and the larger is \(72\). Find the integers.
• The sum of \(3\) times a larger integer and \(2\) times a smaller is \(15\). When \(3\) times the smaller integer is subtracted from twice the larger, the result is \(23\). Find the integers.
• The sum of twice a larger integer and \(3\) times a smaller is \(10\). When the \(4\) times the smaller integer is added to the larger, the result is \(0\). Find the integers.
• The difference of twice a smaller integer and \(7\) times a larger is \(4\). When \(5\) times the larger integer is subtracted from \(3\) times the smaller, the result is \(−5\). Find the integers.
• The difference of a smaller integer and twice a larger is \(0\). When \(3\) times the larger integer is subtracted from \(2\) times the smaller, the result is \(−5\). Find the integers.
• The length of a rectangle is \(5\) more than twice its width. If the perimeter measures \(46\) meters, then find the dimensions of the rectangle.
• The width of a rectangle is \(2\) centimeters less than one-half its length. If the perimeter measures \(62\) centimeters, then find the dimensions of the rectangle.
• A partitioned rectangular pen next to a river is constructed with a total \(136\) feet of fencing (see illustration). If the outer fencing measures \(114\) feet, then find the dimensions of the pen.

14. A partitioned rectangular pen is constructed with a total \(168\) feet of fencing (see illustration). If the perimeter measures \(138\) feet, then find the dimensions of the pen.

15. Find \(a\) and \(b\) such that the system \(\left\{ \begin{array} { l } { a x + b y = 8 } \\ { b x + a y = 7 } \end{array} \right.\) has solution \((2,1)\). (Hint: Substitute the given \(x\)- and \(y\)-values and solve the resulting linear system in terms of \(a\) and \(b\).)

16. Find \(a\) and \(b\) such that the system \(\left\{ \begin{array} { l } { a x - b y = 11 } \\ { b x + a y = 13 } \end{array} \right.\) has solution \((3, -1)\).

17. A line passes through two points \((5, −9)\) and \((−3, 7)\). Use these points and \(y = mx + b\) to construct a system of two linear equations in terms of \(m\) and \(b\) and solve it.

18. A line passes through two points \((2, 7)\) and \((\frac{1}{2}, −2)\). Use these points and \(y = mx + b\) to construct a system of two linear equations in terms of \(m\) and \(b\) and solve it.

19. A \($5,200\) principal is invested in two accounts, one earning \(3\)% interest and another earning \(6\)% interest. If the total interest for the year is \($210\), then how much is invested in each account?

20. Harry’s \($2,200\) savings is in two accounts. One account earns \(2\)% annual interest and the other earns \(4\)%. His total interest for the year is \($69\). How much does he have in each account?

21. Janine has two savings accounts totaling \($6,500\). One account earns \(2 \frac{3}{4}\)% annual interest and the other earns \(3 \frac{1}{2}\)%. If her total interest for the year is \($211\), then how much is in each account?

22. Margaret has her total savings of \($24,200\) in two different CD accounts. One CD earns \(4.6\)% interest and another earns \(3.4\)% interest. If her total interest for the year is \($1,007.60\), then how much does she have in each CD account?

23. Last year Mandy earned twice as much interest in her Money Market fund as she did in her regular savings account. The total interest from the two accounts was \($246\). How much interest did she earn in each account?

24. A small business invested \($120,000\) in two accounts. The account earning \(4\)% annual interest yielded twice as much interest as the account earning \(3\)% annual interest. How much was invested in each account?

25. Sally earns \($1,000\) per month plus a commission of \(2\)% of sales. Jane earns \($200\) per month plus \(6\)% of her sales. At what monthly sales figure will both Sally and Jane earn the same amount of pay?

26. The cost of producing specialty book shelves includes an initial set-up fee of \($1,200\) plus an additional \($20\) per unit produced. Each shelf can be sold for \($60\) per unit. Find the number of units that must be produced and sold where the costs equal the revenue generated.

27. Jim was able to purchase a pizza for \($12.35\) with quarters and dimes. If he uses \(71\) coins to buy the pizza, then how many of each did he have?

28. A cash register contains \($5\) bills and \($10\) bills with a total value of \($350\). If there are \(46\) bills total, then how many of each does the register contain?

29. Two families bought tickets for the home basketball game. One family ordered \(2\) adult tickets and \(4\) children’s tickets for a total of \($36.00\). Another family ordered \(3\) adult tickets and \(2\) children’s tickets for a total of \($32.00\). How much did each ticket cost?

30. Two friends found shirts and shorts on sale at a flea market. One bought \(4\) shirts and \(2\) shorts for a total of \($28.00\). The other bought \(3\) shirts and \(3\) shorts for a total of \($30.75\). How much was each shirt and each pair of shorts?

31. A community theater sold \(140\) tickets to the evening musical for a total of \($1,540\). Each adult ticket was sold for \($12\) and each child ticket was sold for \($8\). How many adult tickets were sold?

32. The campus bookstore sells graphing calculators for \($110\) and scientific calculators for \($16\). On the first day of classes \(50\) calculators were sold for a total of \($1,646\). How many of each were sold?

33. A jar consisting of only nickels and quarters contains \(70\) coins. If the total value is \($9.10\), how many of each coin are in the jar?

34. Jill has \($9.20\) worth of dimes and quarters. If there are \(68\) coins in total, how many of each does she have?

1. The integers are \(16\) and \(29\).

3. The integers are \(6\) and \(35\).

5. The integers are \(25\) and \(36\).

7. The integers are \(−3\) and \(7\).

9. The integers are \(−5\) and \(−2\).

11. Length: \(17\) meters; width: \(6\) meters

13. Width: \(22\) feet; length: \(70\) feet

15. \(a = 3, b = 2\)

17. \(m = −2, b = 1\)

19. \($3,400\) at \(3\)% and \($1,800\) at \(6\)%

21. \($2,200\) at \(2 \frac{3}{4}\)% and \($4,300\) at \(3 \frac{1}{2}\)%

23. Savings: \($82\); Money Market: \($164\).

25. \($20,000\)

27. \(35\) quarters and \(36\) dimes

29. Adults \($7.00\) each and children \($5.50\) each.

31. \(105\) adult tickets were sold.

33. The jar contains \(42\) nickels and \(28\) quarters.

Exercise \(\PageIndex{6}\)

• A \(17\)% acid solution is to be mixed with a \(9\)% acid solution to produce \(8\) gallons of a \(10\)% acid solution. How much of each is needed?
• A nurse wishes to obtain \(28\) ounces of a \(1.5\)% saline solution. How much of a \(1\)% saline solution must she mix with a \(4.5\)% saline solution to achieve the desired mixture?
• A customer ordered \(4\) pounds of a mixed peanut product containing \(12\)% cashews. The inventory consists of only two mixes containing \(10\)% and \(26\)% cashews. How much of each type must be mixed to fill the order?
• One alcohol solution contains \(10\)% alcohol and another contains \(25\)% alcohol. How much of each should be mixed together to obtain \(2\) gallons of a \(13.75\)% alcohol solution?
• How much cleaning fluid concentrate, with \(60\)% alcohol content, must be mixed with water to obtain a \(24\)-ounce mixture with \(15\)% alcohol content?
• How many pounds of pure peanuts must be combined with a \(20\)% peanut mix to produce \(2\) pounds of a \(50\)% peanut mix?
• A \(50\)% fruit juice concentrate can be purchased wholesale. Best taste is achieved when water is mixed with the concentrate in such a way as to obtain a \(15\)% fruit juice mixture. How much water and concentrate is needed to make a \(60\)-ounce fruit juice drink?
• Pure sugar is to be mixed with a fruit salad containing \(10\)% sugar to produce \(65\) ounces of a salad containing \(18\)% sugar. How much pure sugar is required?
• A custom aluminum alloy is created by mixing \(150\) grams of a \(15\)% aluminum alloy and \(350\) grams of a \(55\)% aluminum alloy. What percentage of aluminum is in the resulting mixture?
• A research assistant mixed \(500\) milliliters of a solution that contained a \(12\)% acid with \(300\) milliliters of water. What percentage of acid is in the resulting solution?

1. \(7\) gallons of the \(9\)% acid solution and \(1\) gallon of the \(17\)% acid solution

3. \(3.5\) pounds of the \(10\)% cashew mix and \(0.5\) pounds of the \(26\)% cashew mix

5. \(6\) ounces of cleaning fluid concentrate

7. \(18\) ounces of fruit juice concentrate and \(42\) ounces of water

Exercise \(\PageIndex{7}\)

• The two legs of a \(432\)-mile trip took \(8\) hours. The average speed for the first leg of the trip was \(52\) miles per hour and the average speed for the second leg of the trip was \(60\) miles per hour. How long did each leg of the trip take?
• Jerry took two buses on the \(265\)-mile trip from Los Angeles to Las Vegas. The first bus averaged \(55\) miles per hour and the second bus was able to average \(50\) miles per hour. If the total trip took \(5\) hours, then how long was spent in each bus?
• An executive was able to average \(48\) miles per hour to the airport in her car and then board an airplane that averaged \(210\) miles per hour. The \(549\)-mile business trip took \(3\) hours. How long did it take her to drive to the airport?
• Joe spends \(1\) hour each morning exercising by jogging and then cycling for a total of \(15\) miles. He is able to average \(6\) miles per hour jogging and \(18\) miles per hour cycling. How long does he spend jogging each morning?
• Swimming with the current Jack can swim \(2.5\) miles in \(\frac{1}{2}\) hour. Swimming back, against the same current, he can only swim \(2\) miles in the same amount of time. How fast is the current?
• A light aircraft flying with the wind can travel \(180\) miles in \(1 \frac{1}{2}\) hours. The aircraft can fly the same distance against the wind in \(2\) hours. Find the speed of the wind.
• A light airplane flying with the wind can travel \(600\) miles in \(4\) hours. On the return trip, against the wind, it will take \(5\) hours. What are the speeds of the airplane and of the wind?
• A boat can travel \(15\) miles with the current downstream in \(1 \frac{1}{4}\) hours. Returning upstream against the current, the boat can only travel \(8 \frac{3}{4}\) miles in the same amount of time. Find the speed of the current.
• Mary jogged the trail from her car to the cabin at the rate of \(6\) miles per hour. She then walked back to her car at a rate of \(4\) miles per hour. If the entire trip took \(1\) hour, then how long did it take her to walk back to her car?
• Two trains leave the station traveling in opposite directions. One train is \(8\) miles per hour faster than the other and in \(2 \frac{1}{2}\) hours they are \(230\) miles apart. Determine the average speed of each train.
• Two trains leave the station traveling in opposite directions. One train is \(12\) miles per hour faster than the other and in \(3\) hours they are \(300\) miles apart. Determine the average speed of each train.
• A jogger can sustain an average running rate of \(8\) miles per hour to his destination and \(6\) miles an hour on the return trip. Find the total distance the jogger ran if the total time running was \(1 \frac{3}{4}\) hour.

1. The first leg of the trip took \(6\) hours and the second leg took \(2\) hours.

3. It took her \(\frac{1}{2}\) hour to drive to the airport.

5. \(0.5\) miles per hour.

7. Airplane: \(135\) miles per hour; wind: \(15\) miles per hour

9. \(\frac{3}{5}\) hour

11. One train averaged \(44\) miles per hour and the other averaged \(56\) miles per hour.

Exercise \(\PageIndex{8}\)

• Compose a number or money problem of your own and share it on the discussion board.
• Compose a mixture problem of your own and share it on the discussion board.
• Compose a uniform motion problem of your own and share it on the discussion board.

1. Answer may vary

3. Answer may vary

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!

Two-Variable Word Problems

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• Class 10 Maths

Important Questions Class 10 Maths Chapter 3 Linear Equations In Two Variables

Important questions of Chapter 3 Linear Equations In Two Variables with solutions for class 10 Maths are available for students preparing for the CBSE board exam 2022-2023. By practising these questions, students will score good marks, covering the major topics included in the NCERT syllabus. These questions are created by subject experts after thorough research and are based on the latest exam pattern. Students can expect a few questions from these important questions in the exam. So, they are advised to practice them well to score high marks in the exam.

Check: Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables MCQs

Chapter 3 linear equations In two variables deals with finding the solutions for pairs of such equations with two variables present. These solutions can also be represented in graphs. Let us solve the questions which are important for the final exams of the 10th standard. Students can practice the additional questions provided at the end of the page for a better understanding of the concept and score maximum marks in the board exam.

Also Check:

• Important 2 Marks Questions for CBSE 10th Maths
• Important 3 Marks Questions for CBSE 10th Maths
• Important 4 Marks Questions for CBSE 10th Maths

Class 10 Maths Chapter 3 Important Questions With Answers

Q.1: The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically.

Let the cost of 1 kg of apples be ‘Rs. x’. And, let the cost of 1 kg of grapes be ‘Rs. y’. According to the question, the algebraic representation is 2x + y = 160 And 4x + 2y = 300

For, 2x + y = 160 or y = 160 − 2x, the solution table is;

For 4x + 2y = 300 or y = (300 – 4x)/ 2, the solution table is;

Note: Students can also represent these two equations graphically, by using the given points of x-coordinate and y-coordinate.

Q.2: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 ……….(1) Given, the length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36 2x + 4 = 36 2x = 32 x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Q.3: On comparing the ratios a 1 /a 2 , b 1 /b 2 , and c 1 /c 2 , find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0 and 2x – 3y = 7 or 2x – 3y – 7 = 0

Comparing the above equations with a 1 x + b 1 y + c 1 =0 And a 2 x + b 2 y + c 2 = 0 We get, a 1  = 3, b 1  = 2, c 1  = -5 a 2  = 2, b 2  = -3, c 2  = -7

a 1/ a 2  = 3/2, b 1 /b 2  = 2/-3, c 1 /c 2  = -5/-7 = 5/7 Since, a 1 /a 2 ≠b 1 /b 2 the lines intersect each other at a point and have only one possible solution.

Hence, the equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9 Therefore, a 1  = 2, b 1  = -3, c 1  = -8 a 2  = 4, b 2  = -6, c 2  = -9

a 1 /a 2  = 2/4 = 1/2, b 1 /b 2  = -3/-6 = 1/2, c 1 /c 2  = -8/-9 = 8/9

Since, a 1 /a 2 =b 1 / b 2 ≠c 1 /c 2

Therefore, the lines are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

Q.4: Solve the following pair of linear equations by the substitution method.

(i) x + y = 14 x – y = 4

(ii) 3x – y = 3 9x – 3y = 9

(i) Given, x + y = 14 and x – y = 4 are the two equations. From 1st equation, we get, x = 14 – y Now, put the value of x in second equation to get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5 By the value of y, we can now find the value of x; ∵ x = 14 – y ∴ x = 14 – 5 Or x = 9 Hence, x = 9 and y = 5.

(ii) Given, 3x – y = 3 and 9x – 3y = 9 are the two equations. From 1st equation, we get, x = (3 + y)/3 Now, substitute the value of x in the given second equation to get, 9[(3 + y)/3] – 3y = 9

⇒ 3(3+y) – 3y = 9 ⇒ 9 + 3y – 3y = 9 ⇒ 9 = 9 Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

Q.5: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

2x + 3y = 11…………………………..(i) 2x – 4y = -24………………………… (ii) From equation (i), we get; x = (11 – 3y)/2 ……….…………………………..(iii)

Putting the value of x in equation (ii), we get

2[(11 – 3y)/2] – 4y = −24 11 – 3y – 4y = -24 -7y = -35 y = 5……………………………………..(iv) Putting the value of y in equation (iii), we get;

x = (11 – 15)/2 = -4/2 = −2 Hence, x = -2, y = 5 Also, y = mx + 3 5 = -2m +3 -2m = 2 m = -1 Therefore, the value of m is -1.

Q.6: The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball. Solution: Let the cost of a bat be x and the cost of a ball be y. According to the question, 7x + 6y = 3800 ………………. (i) 3x + 5y = 1750 ………………. (ii) From (i), we get; y = (3800 – 7x)/6 …………………… (iii) Substituting (iii) in (ii). we get, 3x + 5[(3800 – 7x)/6] = 1750 ⇒3x + (9500/3) – (35x/6) = 1750 3x – (35x/6) = 1750 – (9500/3)

(18x – 35x)/6 = (5250 – 9500)/3 ⇒-17x/6 = -4250/3 ⇒-17x = -8500 x = 500 Putting the value of x in (iii), we get;

y = (3800 – 7 × 500)/6 = 300/6 = 50 Hence, the cost of a bat is Rs 500 and the cost of a ball is Rs 50.

Q.7: A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution: Let the fraction be x/y.

According to the question, (x + 2)/(y + 2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1) (x + 3)/(y + 3) = 5/6

6x + 18 = 5y +15 6x – 5y = -3 ………………. (2) From (1), we get

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2), we get 6[(-4 + 9y)/11] – 5y = -3

-24 + 54y – 55y = -33 -y = -9 y = 9 ………………… (4) Substituting the value of y in (3), we get x = (-4 + 81)/11 = 77/11 = 7

Hence, the fraction is 7/9.

Q.8 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? Solution: Let us assume, the present age of Nuri be x. And the present age of Sonu is y. According to the given condition, we can write as; x – 5 = 3(y – 5) x – 3y = -10…………………………………..(1) Now, x + 10 = 2(y +10) x – 2y = 10…………………………………….(2) Subtract eq. 1 from 2, to get, y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get, x – 3(20) = -10 x – 60 = -10 x = 50 Therefore, The age of Nuri is 50 years The age of Sonu is 20 years.

(ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Solution:

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B. According to the information given, A + 4B = 27 …………………………………….…………………………. (i) A + 2B = 21 ……………………………………………………………….. (ii) When equation (ii) is subtracted from equation (i) we get, 2B = 6 B = 3 …………………………………………………………………………(iii) Substituting B = 3 in equation (i) we get, A + 12 = 27 A = 15 Hence, the fixed charge is Rs. 15. And the Additional charge per day is Rs. 3.

Q.9: Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9 3x + 2y = 4

8x + 5y = 9 …………………..(1) 3x + 2y = 4 ……………….….(2) From equation (2) we get;

x = (4 – 2y) / 3 ……………………. (3) Substituting this value in equation 1, we get 8[(4 – 2y)/3] + 5y = 9 32 – 16y + 15y = 27 -y = -5 y = 5 ……………………………….(4) Substituting this value in equation (2), we get 3x + 10 = 4

3x = -6 x = -2 Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method: 8x +5y – 9 = 0 3x + 2y – 4 = 0

x/(-20 + 18) = y/(-27 + 32 ) = 1/(16 – 15) -x/2 = y/5 = 1/1 ∴ x = -2 and y =5.

Q.10: Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(i) Let us consider, Speed of boat is still water = x km/hr Speed of current = y km/hr Now, speed of Ritu, during, Downstream = x + y km/hr Upstream = x – y km/hr As per the question given, 2(x + y) = 20 Or x + y = 10……………………….(1) And, 2(x – y) = 4 Or x – y = 2………………………(2) Adding both the eq.1 and 2, we get, 2x = 12 x = 6 Putting the value of x in eq.1, we get, y = 4 Therefore,

Speed of Ritu is still water = 6 km/hr Speed of current = 4 km/hr

Q.11: Solve the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0 graphically.

x + 2y – 4 = 0….(i)

2x + 4y – 12 = 0….(ii)

2y = 4 – x

y = (4 – x)/2

2x + 4y = 12

2y = 6 – x

y = (6 – x)/2

Plotting the points on the graph, we get;

Here, the lines represent the given pair of linear equations are parallel.

Thus, there is no solution to the given pair of linear equations.

Q.12: Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution.

3x + ky + 15 = 0

Also, given that the pair of equations has a unique solution.

Comparing the given equations with standard form,

a 1 = 1, b 1 = 2, c 1 = -5

a 2 = 3, b 2 = k, c 2 = 15

Condition for unique solution is:

a 1 /a 2 ≠ b 1 /b 2

Thus, for all real values of k except 6, the given pair of equations has a unique solution.

Q.13: Determine graphically the coordinates of vertices of a triangle, the equation of whose sides are given by 2y – x = 8, 5y – x = 14 and y – 2x = 1.

2y – x = 8….(i)

5y – x = 14….(ii)

y – 2x = 1….(iii)

y = (x + 8)/2

5y = x + 14

y = (x + 14)/5

From (iii),

Let us plot all these points on the graph.

From the graph, we can write the coordinates of vertices of triangle formed are:

P(-4, 2), Q(1, 3), and R(2, 5)

Q.14: Use elimination method to find all possible solutions of the following pair of linear equation:

2x + 3y = 8 

4x + 6y = 7

2x + 3y = 8….(i)

4x + 6y = 7….(ii)

Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal.

4x + 6y = 16….(iii)

4x + 6y = 7….(iv)

Subtracting (iv) from (iii),

4x + 6y – 4x – 6y = 16 – 7

0 = 9, it is not possible

Therefore, the pair of equations has no solution.

Q.15: Solve the following pairs of equations by reducing them to a pair of linear equations:

1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let us assume 1/x = m and 1/y = n , then the equations will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

Pair of Linear Equations in Two Variables Questions for Practice

• A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
• For which values of a and b do the following pair of linear equations have infinite solutions? 2x + 3y = 7 (a – b) x + (a + b) y = 3a + b – 2
• Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
• Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
• Two lines are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.
• Solve the following pair of equations by substitution method: 7x – 15y = 2 x + 2y = 3
• The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.
• Solve the following pair of linear equations by the substitution and cross-multiplication methods : 8x + 5y = 9 3x + 2y = 4
• Solve the following pairs of equations by reducing them to a pair of linear equations: 6x + 3y = 6xy 2x + 4y = 5xy
• The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Students must have found these important questions of Class 10 Maths Chapter 3 Linear Equations In Two Variables helpful for the exam preparation. Keep practising and stay tuned to BYJU’S for the latest update on CBSE/ICSE/State Board/Competitive exams.

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Your extra questions really helped me 4 d preparation for mid-term examination. ………..Tysm

NICE QUESTIONS

it is exellent for mid term and class tests practice

It was very helpful for me for preparation for my examinations Ty👍

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SOLVING SYSTEM OF LINEAR EQUATIONS WORD PROBLEMS IN 2 VARIABLES

Problem 1 :

A fraction becomes 9/11, if 2 is added to both numerator and the denominator. If 3 be added to both the numerator and the denominator, the fraction becomes 5/6. Find the fraction.

Let x/y be the required fraction

If 2 is added to both numerator and denominator the fraction becomes 9/11.

(x + 2)/(y+2)  =  9/11

11(x + 2)  =  9(y + 2)

11x + 22  =  9y + 18

11x – 9y  =  18 – 22

11x – 9y  =  -4 ------(1)

If 3 is added to both the numerator and the denominator it becomes 5/6.

(x + 3)/(y+3)  =  5/6

6(x + 3)  =  5(y + 3)

6x + 18  =  5y + 15

6x – 5y  =  15 - 18

6x – 5y  =  - 3 ------(2)

Solving (1) and (2), we get

                                               x  =  7

Substitute 7 for x in (1). 

(1)-----> 11(7) – 9y  =  -4

77 – 9y  =  -4

-9y  =  -81

y  =  9

Then, 

x/y  =  7/9

So, the required fraction is 7/9.

Problem 2 :

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Let x be the age of Jacob and y be the age of his son.

5 years hence their ages will be

x + 5 and y + 5

x + 5  =  3(y + 5)

x + 5  =  3y + 15

 x – 3y  =  -5 + 15

 x – 3y  =  10 ------(1)

5 years ago, their ages were

x – 5 and y – 5

x – 5  =  7(y – 5)

x – 5  =  7y – 35

x – 7y  =  5 - 35

x – 7y  =  -30 ------(2)

(1) - (2) :

4y  =  40

y  =  10

Substitute 10 for y in (2).

(2)------> x - 7(10)  =  -30

x – 70  =  -30

x  =  40

age of Jacob's son  =  10

Age of Jacob  =  40

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