case study questions of linear equations in two variables

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CBSE Class 10 Maths Case Study Questions for Chapter 3 - Pair of Linear Equations in Two Variables (Published by CBSE)

Cbse's question bank on case study for class 10 maths chapter 3 is available here. these questions will be very helpful to prepare for the cbse class 10 maths exam 2022..

Case study questions are going to be new for CBSE Class 10 students. These are the competency-based questions that are completely new to class 10 students. To help students understand the format of the questions, CBSE has released a question bank on case study for class 10 Maths. Students must practice with these questions to get familiarised with the concepts and logic used in the case study and understand how to answers them correctly. You may check below the case study questions for CBSE Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables. You can also check the right answer at the end of each question.

Check Case Study Questions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables

CASE STUDY-1:

1. If answer to all questions he attempted by guessing were wrong, then how many questions did he answer correctly?

2. How many questions did he guess?

3. If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?

4. If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks?

Let the no of questions whose answer is known to the student x and questions attempted by cheating be y

x – 1/4y =90

solving these two

x = 96 and y = 24

1. He answered 96 questions correctly.

2. He attempted 24 questions by guessing.

3. Marks = 80- ¼ 0f 40 =70

4. x – 1/4 of (120 – x) = 95

5x = 500, x = 100

CASE STUDY-2:

Amit is planning to buy a house and the layout is given below. The design and the measurement has been made such that areas of two bedrooms and kitchen together is 95 sq.m.

Based on the above information, answer the following questions:

1. Form the pair of linear equations in two variables from this situation.

2. Find the length of the outer boundary of the layout.

3. Find the area of each bedroom and kitchen in the layout.

4. Find the area of living room in the layout.

5. Find the cost of laying tiles in kitchen at the rate of Rs. 50 per sq.m.

1. Area of two bedrooms= 10x sq.m

Area of kitchen = 5y sq.m

10x + 5y = 95

Also, x + 2+ y = 15

2. Length of outer boundary = 12 + 15 + 12 + 15 = 54m

3. On solving two equation part(i)

x = 6m and y = 7m

area of bedroom = 5 x 6 = 30m

area of kitchen = 5 x 7 = 35m

4. Area of living room = (15 x 7) – 30 = 105 – 30 = 75 sq.m

5. Total cost of laying tiles in the kitchen = Rs50 x 35 = Rs1750

Case study-3 :

It is common that Governments revise travel fares from time to time based on various factors such as inflation ( a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, Rickshaws, taxis, Radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations:

Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110.

Situation 2: In a city B, for a journey of 8km, the charge paid is Rs91 and for a journey of 14km, the charge paid is Rs 145.

Refer situation 1

1. If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr, the pair of linear equations representing the situation is

a) x + 10y =110, x + 15y = 75

b) x + 10y = 75, x + 15y = 110

c) 10x + y = 110, 15x + y = 75

d) 10x + y = 75, 15x + y = 110

Answer: b) x + 10y = 75, x + 15y = 110

2. A person travels a distance of 50km. The amount he has to pay is

Answer: c) Rs.355

Refer situation 2

3. What will a person have to pay for travelling a distance of 30km?

Answer: b) Rs.289

4. The graph of lines representing the conditions are: (situation 2)

Answer: (iii)

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

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Class 10 Maths Case Study Questions Chapter 3 Pair of Linear Equations in Two Variables

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Case study Questions in the Class 10 Mathematics Chapter 3  are very important to solve for your exam. Class 10 Maths Chapter 3 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 3  Pair of Linear Equations in Two Variables

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Pair of Linear Equations in Two Variables Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 3 Pair of Linear Equations in Two Variables

Case Study/Passage-Based Questions

(i) 1 st  situation can be represented algebraically as

Answer: (d) 2x+3y=46

(ii) 2 nd  situation can be represented algebraically as

Answer: (c) 3x + 5y = 74

(iii), Fare from Ben~aluru to Malleswaram is

Answer: (b) Rs 8

(iv) Fare from Bengaluru to Yeswanthpur is

Answer: (a) Rs 10

(v) The system oflinear equations represented by both situations has

Answer: (c) unique solution

Case Study 2: The scissors which are so common in our daily life use, its blades represent the graph of linear equations.

Let the blades of a scissor are represented by the system of linear equations:

x + 3y = 6 and 2x – 3y = 12

(i) The pivot point (point of intersection) of the blades represented by the linear equation x + 3y = 6 and 2x – 3y = 12 of the scissor is (a) (2, 3) (b) (6, 0) (c) (3, 2) (d) (2, 6)

Answer: (b) (6, 0)

(ii) The points at which linear equations x + 3y = 6 and 2x – 3y = 12 intersect y – axis respectively are (a) (0, 2) and (0, 6) (b) (0, 2) and (6, 0) (c) (0, 2) and (0, –4) (d) (2, 0) and (0, –4)

Answer: (c) (0, 2) and (0, –4)

(iii) The number of solution of the system of linear equations x + 2y – 8 = 0 and 2x + 4y = 16 is (a) 0 (b) 1 (c) 2 (d) infinitely many

Answer: (d) infinitely many

(iv) If (1, 2) is the solution of linear equations ax + y = 3 and 2x + by = 12, then values of a and b are respectively (a) 1, 5 (b) 2, 3 (c) –1, 5 (d) 3, 5

Answer: (a) 1, 5

(v) If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (a) intersecting (b) parallel (c) always coincident (d) intersecting or coincident

Answer: (d) intersecting or coincident

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Pair of Linear Equations in Two Variables Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Tuesday, September 28, 2021

Class 10 maths case study based questions chapter 3 pair of linear equations cbse board term 1 with answer key.

          Hello students, Welcome to Maths Easy Institute.  

CASE STUDY 1:   

A library is a collection of materials, books, and media that are easily accessible to everyone. Here, you can find books from different genres such that as science fiction, fiction, and many research papers. It is also a great place to socialize with your community members, which will help you build relationships with people of similar interests.

One day, two friends Sarita and Babita go to a library for some books. The library has fixed charges for the first 3 days and an additional charge for each day thereafter. If anyone takes the book for 10 days then he/she has to pay fixed charges for 3 days and additional charges for 7 days. 

Sarita take a book from Library for 7 days and paid 27 rs and Babita  take a book from Library for 5 days and paid 21 rs .

(a) If fixed charges for the library for the first 3 days is x Rs and additional charges for each day is y Rs/day then pair of linear equation satisfying the  Sarita case is:

(b)  If fixed charges for the library for first 3 days is x Rs and additional charges for each is y Rs/day then pair of linear equation satisfying Babita case is:

(c) Fixed charges and additional charges of the library is:

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Case Study Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

• Last modified on: 10 months ago
• Reading Time: 8 Minutes

Case Study Questions

Question 1:

Anil went to buy some vegetables, he bought ‘x’ kgs. of tomato and ‘y’ kgs. of potato. The total cost of vegetables comes out to be of Rs. 200. Now if the cost of 1 kg of tomato is Rs. 50 and 1 kg of potato is Rs. 20, then answer the following questions.

(i) Which of the following equations represent the total cost. (a) 5x – 2y = 20 (b) 5y + 2x = 20 (c) 5x + 2y = 20 (d) 2x + 5y = 20

(ii) If Anil bought ‘x’ kgs of tomato and 2.5 kgs. of potato, then find the value of ‘x’. (a) 5 (b) 2 (c) 3 (d) 4

(iii) If Anil bought ‘2’ kgs of tomato and ‘y’ kgs of potato, then find the value of ‘y’. (a) 5 (b) 2 (c) 3 (d) 4

(iv) The graph of 5x + 2y = 20 cuts x-axis at the point. (a) (10, 0) (b) (4, 0) (c) (0, 0) (d) it is parallel to x-axis

(v) The graph of 5x + 2y = 20 cuts y-axis at the point. (a) (0, 10) (b) (0, 4) (c) (0, 0) (d) it is parallel to y-axis

(i) C (ii) C (iii) A (iv) B (v) A

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4 thoughts on “ Case Study Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables ”

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Answer key (i)(c) 5x + 2y = 20 (ii)(c) 3 (iii)(a) 5 (iv)(b) (4, 0) (v)(a) (0, 10)

Hlo sir I have doubt

kindly solve ii and iii part

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CBSE Class 10 Maths: Case Study Questions of Chapter 3 Pair of Linear Equations in Two Variables PDF Download

Case study Questions in the Class 10 Mathematics Chapter 3  are very important to solve for your exam. Class 10 Maths Chapter 3 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 3  Pair of Linear Equations in Two Variables

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Pair of Linear Equations in Two Variables Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 3 Pair of Linear Equations in Two Variables

Case Study/Passage-Based Questions

Question 1:

(i) 1 st  situation can be represented algebraically as

Answer: (d) 2x+3y=46

(ii) 2 nd  situation can be represented algebraically as

Answer: (c) 3x + 5y = 74

(iii), Fare from Ben~aluru to Malleswaram is

Answer: (b) Rs 8

(iv) Fare from Bengaluru to Yeswanthpur is

Answer: (a) Rs 10

(v) The system oflinear equations represented by both situations has

Answer: (c) unique solution

Question 2:

The scissors which is so common in our daily life use, its blades represent the graph of linear equations.

Let the blades of a scissor are represented by the system of linear equations:

x + 3y = 6 and 2x – 3y = 12

(i) The pivot point (point of intersection) of the blades represented by the linear equation x + 3y = 6 and 2x – 3y = 12 of the scissor is (a) (2, 3) (b) (6, 0) (c) (3, 2) (d) (2, 6)

Answer: (b) (6, 0)

(ii) The points at which linear equations x + 3y = 6 and 2x – 3y = 12 intersect y – axis respectively are (a) (0, 2) and (0, 6) (b) (0, 2) and (6, 0) (c) (0, 2) and (0, –4) (d) (2, 0) and (0, –4)

Answer: (c) (0, 2) and (0, –4)

(iii) The number of solution of the system of linear equations x + 2y – 8 = 0 and 2x + 4y = 16 is (a) 0 (b) 1 (c) 2 (d) infinitely many

Answer: (d) infinitely many

(iv) If (1, 2) is the solution of linear equations ax + y = 3 and 2x + by = 12, then values of a and b are respectively (a) 1, 5 (b) 2, 3 (c) –1, 5 (d) 3, 5

Answer: (a) 1, 5

(v) If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (a) intersecting (b) parallel (c) always coincident (d) intersecting or coincident

Answer: (d) intersecting or coincident

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Pair of Linear Equations in Two Variables Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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4.1 Solve Systems of Linear Equations with Two Variables

Learning objectives.

By the end of this section, you will be able to:

• Determine whether an ordered pair is a solution of a system of equations
• Solve a system of linear equations by graphing
• Solve a system of equations by substitution
• Solve a system of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Be Prepared 4.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 , y = 2 3 x − 4 , ⓐ Is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ Is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 3.2 .

Be Prepared 4.2

Find the slope and y -intercept of the line 3 x − y = 12 . 3 x − y = 12 . If you missed this problem, review Example 3.16 .

Be Prepared 4.3

Find the x- and y -intercepts of the line 2 x − 3 y = 12 . 2 x − 3 y = 12 . If you missed this problem, review Example 3.8 .

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

A linear equation in two variables, such as 2 x + y = 7 , 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ) . ( x , y ) .

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example 4.1

Determine whether the ordered pair is a solution to the system { x − y = −1 2 x − y = −5 . { x − y = −1 2 x − y = −5 .

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

Determine whether the ordered pair is a solution to the system { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system { x − 3 y = −8 − 3 x − y = 4 . { x − 3 y = −8 − 3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

• Step 1. Graph the first equation.
• Step 2. Graph the second equation on the same rectangular coordinate system.
• Step 3. Determine whether the lines intersect, are parallel, or are the same line.
• If the lines intersect, identify the point of intersection. This is the solution to the system.
• If the lines are parallel, the system has no solution.
• If the lines are the same, the system has an infinite number of solutions.
• Step 5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts.

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Solve the system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Example 4.5

Solve the system by graphing: { y = 2 x − 3 − 6 x + 3 y = − 9 . { y = 2 x − 3 − 6 x + 3 y = − 9 .

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident Lines

Coincident lines have the same slope and same y- intercept.

The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.

In Example 4.5 , the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4.4 , has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1 .

Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ { y = 3 x − 1 6 x − 2 y = 12 { y = 3 x − 1 6 x − 2 y = 12 ⓑ { 2 x + y = − 3 x − 5 y = 5 { 2 x + y = − 3 x − 5 y = 5

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Try It 4.11

ⓐ { y = −2 x − 4 4 x + 2 y = 9 { y = −2 x − 4 4 x + 2 y = 9 ⓑ { 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 4.12

ⓐ { y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6 ⓑ { x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

Solve a system of equations by substitution.

• Step 1. Solve one of the equations for either variable.
• Step 2. Substitute the expression from Step 1 into the other equation.
• Step 3. Solve the resulting equation.
• Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
• Step 5. Write the solution as an ordered pair.
• Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .

Then rewrite the system of equations.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 4.9

How to solve a system of equations by elimination.

Solve the system by elimination: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

• Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Step 3. Add the equations resulting from Step 2 to eliminate one variable.
• Step 4. Solve for the remaining variable.
• Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
• Step 6. Write the solution as an ordered pair.
• Step 7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ − 2 x + 4 y = 4 y = 1 2 x { − 2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: { 3 x + y = 12 x = y − 8 . { 3 x + y = 12 x = y − 8 .

Solve the system of equations by elimination and explain all your steps in words: { 5 x + 4 y = 10 2 x = 3 y + 27 . { 5 x + 4 y = 10 2 x = 3 y + 27 .

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing ⓑ by substitution ⓒ Which method do you prefer? Why?

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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• Book title: Intermediate Algebra 2e
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• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
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Case Study on Pair of Equations in Two Variables Class 10 Maths PDF

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• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Pair of Equations in Two Variables Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Pair of Equations in Two Variables case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Pair of Equations in Two Variables Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Pair of Equations in Two Variables as well as real-life implications of those learnings too.

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• Basic Formulas of Pair of Equations in Two Variables: One of the most important things to know to solve Case Study Questions on Class 10 Pair of Equations in Two Variables is to learn about the basic formulas or revise them before solving the case-based questions on Pair of Equations in Two Variables.
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1.2: Solving Linear Equations in Two Variables

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Learning Objectives

In this section, you will learn to:

• Show all solutions to an equation by graphing a line.
• Graph a line by finding its intercepts.
• Graph and find equations of vertical and horizontal lines.
• Find the slope of a line.
• Graph the line if a point and the slope are given.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. On a piece of graph paper, plot and label these points: A(1, 4), B(-3, 2), C(2, -5), D(0, -3), E(4, 0).

If you missed any part of this problem, review here . (Note that this will open a different textbook in a new window.)

2. Write the coordinates of each of the points shown in this graph.

3. Solve the equation \(2x-6y=24\) for the given values.

a. Solve for \(x\) when \(y=5\).

b. Solve for \(y\) when \(x=3).

a. \(x=27\)

b. \(y=-3\)

If you missed any part of this problem, review Section 1.1 . (Note that this will open a in a new window.)

4. Simplify each expression without the use of a calculator.

a. \(-4-(-7)\)

b. \(-4+(-7)\)

c. \(-4-7\)

d. \(-4+7\)

Solutions of Equations with Two Variables

Remember that a solution is the value(s) of the variable(s) that make(s) the equation true.

When an equation only has one variable, we can isolate that variable to identify the solution. In most equations with one variable, there exists only one value that can make the equation true. For example, in the equation \(6 = x + 2\), the only solution to this equation is \(x = 4\) since that is the only value of the variable that would make the equation true.

When an equation has two variables however, there exist infinite values for the variables that make the equation true. For example, consider the equation \(y = x + 2\).

• If x=0, then y=2.
• If x=1, then y=3.
• If x=2, then y=4.
• If x=100, then y=102.

There are an infinite number of possibilities that can make the equation \(y = x + 2\) true. Rather than try to list all solutions to the equation (which is impossible), we can show all solutions to the equation by graphing them on a coordinate grid. Each solution is listed as an (x, y) point, as shown here:

• If x=0, then y=2. This is shown as the point (0, 2).
• If x=1, then y=3. This is shown as the point (1, 3).
• If x=2, then y=4. This is shown as the point (2, 4).
• If x=100, then y=102. This is shown as the point (100, 102).

The graph for \(y = x + 2\) would look like the graph below. Note that the points (0, 2), (1, 3), and (2, 4) are included among the points that the graph goes through. If the line were drawn out far enough, it would go through the point (100, 102) as well.

Note that this graph goes through infinite points that are not labeled here. Every point the graph touches would be a solution to the equation \(y = x + 2\) though, including (3, 5), (-2, 0), and even (1.5, 3.5). You are encouraged to test each of these pairs of values into the equation to verify that they do indeed make the equation true!

Graphing a Line from its Equation

Equations whose graphs are straight lines are called linear equations. The following are some examples of linear equations:

\(2 x-3 y=6, \quad 3 x=4 y-7, \quad y=2 x-5, \quad 2 y=3, \quad \text { and } \quad x-2=0\)

A line is completely determined by two points. Therefore, to graph a linear equation we need to find the coordinates of two points. This can be accomplished by choosing an arbitrary value for x or y and then solving for the other variable.

Example \(\PageIndex{1}\)

Graph the line: \(y = 3x + 2\)

We need to find the coordinates of at least two points. We arbitrarily choose x = - 1, x = 0, and x = 1.

• If x = -1, then y = 3(-1) + 2 or -1. Therefore, (-1, -1) is a point on this line.
• If x = 0, then y = 3(0) + 2 or y = 2. Hence the point (0, 2).
• If x = 1, then y = 5, and we get the point (1, 5).

Below, the results are summarized, and the line is graphed.

Example \(\PageIndex{2}\)

Graph the line: \(2x + y = 4\)

Again, we need to find coordinates of at least two points.

We arbitrarily choose x = -1, x = 0, and y = 2.

• If x = -1, then 2(-1) + y = 4 which results in y = 6. Therefore, (-1, 6) is a point on this line.
• If x = 0, then 2(0) + y = 4, which results in y = 4. Hence the point (0, 4).
• If y = 2, then 2x + 2 = 4, which yields x = 1, and gives the point (1, 2).

The table below shows the points, and the line is graphed.

Exercise \(\PageIndex{1}\)

Graph the line: \(y = 2x - 3\).

The points at which a line crosses the coordinate axes are called the intercepts . Since the line is touching one axis at each intercept, we know the value of one of the variables at that point.

Consider this graph below. Even though this graph isn't as "simple" as a linear graph, we can still discuss its intercepts.

The x-intercepts on this graph are identified with points. Even without numbers on this graph, we can still know something about the x-intercepts. What do all 4 of these points have in common? Each of them has a y-value of zero. Or to write out the coordinates of these points, they would each look like (_?_ , 0).

Likewise, we could consider the y-intercept of the same graph:

Even without numbers on this graph, we can still know something about the y-intercept. What is it? We know it has an x-value of zero. Or to write out the coordinate of this point, it would look like (0, _?_).

When graphing a line by plotting two points, using the intercepts is often preferred because they are often relatively easy to find.

• To find the value of the x-intercept, we let y = 0 and solve for x.
• To find the value of the y-intercept, we let x = 0 and solve for y.

Example \(\PageIndex{3}\)

Find the intercepts of the line: \(2x - 3y = 6\), and graph.

To find the x-intercept, let y = 0 in the equation and solve for x.

\[\begin{align*} 2x - 3(0) &= 6 \\[4pt] 2x - 0 &= 6 \\[4pt] 2x &= 6 \\[4pt] x &= 3 \end{align*}\]

Therefore, the x-intercept is the point (3,0).

To find the y-intercept, let x = 0 in the equation and solve for y.

\[\begin{align*} 2(0) - 3y &= 6 \\[4pt] 0 - 3y &= 6 \\[4pt] -3y &= 6 \\[4pt] y &= -2 \end{align*}\]

Therefore, the y-intercept is the point (0, -2).

To graph the line, plot the points for the x-intercept (3,0) and the y-intercept (0, -2), and use them to draw the line.

Exercise \(\PageIndex{2}\)

Find the intercepts of the line: \(4x + 2y = 12\), and graph.

Horizontal and Vertical Lines

When an equation of a line has only one variable, the resulting graph is a horizontal or a vertical line.

• The graph of the line \(x = a\), where \(a\) is a constant, is a vertical line that passes through the point \((a, 0)\). Every point on this line has the x-coordinate equal to a, regardless of the y-coordinate.
• The graph of the line \(y = b\), where \(b\) is a constant, is a horizontal line that passes through the point \((0, b)\). Every point on this line has the y-coordinate equal to b, regardless of the x-coordinate.

Example \(\PageIndex{4}\)

Graph the lines: x = -2 , and y = 3.

The graph of the line x = -2 is a vertical line that has the x-coordinate -2 no matter what the y-coordinate is. The graph is a vertical line passing through point (-2, 0).

The graph of the line y = 3, is a horizontal line that has the y-coordinate 3 regardless of what the x-coordinate is. Therefore, the graph is a horizontal line that passes through point (0, 3).

Note: Most students feel that the coordinates of points must always be integers. This is not true, and in real life situations, not always possible. Do not be intimidated if your points include numbers that are fractions or decimals.

Graphing Using Slope

A graph of a line can also be determined if one point and the "steepness" of the line is known. The number that refers to the steepness or inclination of a line is called the slope of the line.

From previous math courses, many of you remember slope as the "rise over run," or "the vertical change over the horizontal change" and have often seen it expressed as:

\[\frac{\text {rise}}{\text {run}}, \frac{\text {vertical change}}{\text {horizontal change}}, \frac{\Delta y}{\Delta x} \text { etc. } \nonumber\]

We give a precise definition.

Definition: Slope

If (\(x_1\), \(y_1\)) and (\(x_2\), \(y_2\)) are two different points on a line, the slope of the line is

\[\text{slope}=m=\frac{y_2-y_1}{x_2-x_1} \label{slope}\]

Example \(\PageIndex{5}\)

Find the slope of the line passing through points (−2, 3) and (4, −1), and graph the line.

Let (\(x_1\), \(y_1\)) = (−2, 3) and (\(x_2\), \(y_2\)) = (4, −1), then the slope (via Equation \ref{slope}) is

To give the reader a better understanding, both the vertical change, -4, and the horizontal change, 6, are shown in the above figure.

When two points are given, it does not matter which point is denoted as (\(x_1\), \(y_1\)) and which (\(x_2\), \(y_2\)). The value for the slope will be the same.

In Example \(\PageIndex{1}\), if we instead choose (\(x_1\), \(y_1\)) = (4, −1) and (\(x_2\), \(y_2\)) = (−2, 3), then we will get the same value for the slope as we obtained earlier.

The steps involved are as follows.

\[m = \frac{3-(-1)}{-2-4} = \frac{4}{-6} = -\frac{2}{3} \nonumber \]

Example \(\PageIndex{6}\)

Find the slope of the line that passes through the points (−1, −4) and (3, −4) and graph the line.

Let (\(x_1\), \(y_1\)) = (−1, −4) and (\(x_2\), \(y_2\)) = (3, −4), then the slope is

\[ m = \frac{-4-(-4)}{3-(-1)} = \frac{0}{4} = 0 \nonumber \]

Note: The slope of any horizontal line is 0

Example \(\PageIndex{7}\)

Find the slope of the line that passes through the points (2, 3) and (2, −1), and graph the line.

Let (\(x_1\), \(y_1\)) = (2, 3) and (\(x_2\), \(y_2\)) = (2, −1) then the slope is

\[m = \frac{-1-3}{2-2}=\frac{4}{0}=\text{undefined.} \nonumber\]

NOTE: Remember that zero can NEVER be in the denominator of a fraction. When we encounter this scenario, the value is undefined. It does not exist. This is different than having a value of zero.

Note: The slope of any vertical line is undefined.

Exercise \(\PageIndex{3}\)

Find the slope of the line that passes through the points (-2, -6) and (2, 4), and graph the line.

Using the previous four examples, students should observe that:

• if a line rises when going from left to right, the line has a positive slope.
• if a line falls going from left to right, the line has a negative slope.
• if a line is horizontal where the value of \(y\) never changes, the line has a slope of zero.
• if a line is vertical where the value of \(x\) never changes, the line does not have a slope.

Sometimes we are given the slope and a point on the line and need to graph the entire line. In this case, we can use the given point as a "starting point" and use slope as \(\frac{\text{rise}}{\text{run}}\) to graph additional points.

Example \(\PageIndex{8}\)

Graph the line that passes through the point (1, 2) and has slope \(-\frac{3}{4}\).

Slope equals \(\frac{\text{rise}}{\text{run}}\). The fact that the slope is \(\frac{-3}{4}\), means that for every rise of -3 units (fall of 3 units) there is a run of 4. So if from the given point (1, 2) we go down 3 units and go right 4 units, we reach the point (5, -1). The graph is obtained by connecting these two points.

Alternatively, since \(\frac{3}{-4}\) represents the same number, the line can be drawn by starting at the point (1,2) and choosing a rise of 3 units followed by a run of -4 units. So from the point (1, 2), we go up 3 units, and to the left 4, thus reaching the point (-3, 5) which is also on the same line. See figure below.

Remember that mathematically, \(-\frac{3}{4}\), \(\frac{-3}{4}\), and \(\frac{3}{-4}\) are all equivalent to one another.

Example \(\PageIndex{9}\)

Find the slope of the line \(2x + 3y = 6\) by finding two points on the line.

In order to find the slope of this line, we will choose any two points on this line.

Again, the selection of \(x\) and \(y\) intercepts seems to be a good choice. The \(x\)-intercept is (3, 0), and the \(y\)-intercept is (0, 2). Therefore, the slope is

\[m = \frac{2-0}{0-3} =-\frac{2}{3}.\nonumber\]

The graph below shows the line and the \(x\)-intercepts and \(y\)-intercepts:

Example \(\PageIndex{10}\)

Find the slope and the \(y\)-intercept of the line \(y = 3x + 2\).

We again find two points on the line, e.g., (0, 2) and (1, 5). Therefore, the slope is

\[m =\frac{5-2}{1-0} = \frac{3}{1} = 3. \nonumber\]

Look at the slopes and the \(y\)-intercepts of the following lines.

It is no coincidence that when an equation of the line is solved for \(y\), the coefficient of the \(x\) term represents the slope, and the constant term represents the \(y\)-intercept.

In other words, for the line \(y = mx + b\), \(m\) is the slope, and \(b\) is the \(y\)-intercept. We will examine this more in the next section.

Example \(\PageIndex{11}\)

Determine the slope and \(y\)-intercept of the line \(2x + 3y = 6\).

We solve for \(y\):

\begin{align*} &2 x+3 y=6 \nonumber \\ &3 y=-2 x+6 \nonumber \\ &y=(-2 / 3) x+2 \nonumber \end{align*}

The slope = the coefficient of the \(x\) term = − 2/3.

The \(y\)-intercept = the constant term = 2.

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Class 9th Maths - Linear Equations in Two Variables Case Study Questions and Answers 2022 - 2023

By QB365 on 08 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Maths Subject - Linear Equations in Two Variables, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Linear equations in two variables case study questions with answer key.

9th Standard CBSE

Final Semester - June 2015

Mathematics

(ii) Find the length of the outer boundary of the layout.

(iii) The pair of linear equation in two variables formed from the statements are (a) x + y = 13, x + y = 9 (b) 2x + y = 13, x + y = 9 (c) x + y = 13, 2x + y = 9 (d) None of the above (iv) Which is the solution satisfying both the equations formed in (iii)?

(v) Find the area of each bedroom.

(iii) Find the cost of one pen?

(iv) Find the total cost if they will purchase the same type of 15 notebooks and 12 pens.

(v) Find whose estimation is correct in the given statement.

(b) How to represent the above situation in linear equations in two variables ?

(c) If Sita contributed Rs. 76, then how much was contributed by Gita ?

(d) If both contributed equally, then how much is contributed by each?

(e) Which is the standard form of linear equations x = – 5 ?

(ii) Which is the solution of the equations formed in (i)?

(c) If the cost of one notebook is Rs. 15 and cost of one pen is 10, then find the total amount.

(d) If the cost of one notebook is twice the cost of one pen, then find the cost of one pen?

(e) Which is the standard form of linear equations y = 4 ?

(b) If the number of children is 15, then find the number of adults?

(c)  If the number of adults is 12, then find the number of children?

(d) Find the value of b, if x = 5, y = 0 is a solution of the equation 3x + 5y = b.

(e) Which is the standard form of linear equations in two variables: y - x = 5?

(b) If the cost of chocolates A is 5, then find the cost of chocolates B?

(c) Which of the follwing point lies on the line x + y = 7?

(d) The point where the line x + y = 7 intersect y-axis is 

(e) For what value of k, x = 2 and y = -1 is a soluation of x + 3y -k = 0.

*****************************************

Linear equations in two variables case study questions with answer key answer keys.

(i) (b) 10x, 5y Area of one bedroom = 5x sq.m Area of two bedrooms = 10x sq.m Area of kitchen = 5y sq. m (ii) (d) 54 m Length of outer boundary = 12 + 15 + 12 + 15 = 54 m (iii) (d) None of the above Area of two bedrooms = 10x sq.m Area of kitchen = 5y sq. m So, 10x + 5y = 95  2x + y = 19 Also, x + 2 + y = 15 x + y = 13 (iv) (c) x = 6, y = 7 x + y = 6 + 7 = 13 2x + y = 2(6) + 7 = 19 x = 6, y = 7 x + y = 6 + 7 = 13 2x + y = 2(6) + 7 = 19  x = 6, y = 7 (v) (a) 30 sq. m Area of living room = (15 x 7) – 30 = 105 – 30 =75 sq. m

(i) (a) 3x + 2y = 80 and 4x + 3y = 110 Here, the cost of one notebook be Rs. x and that of pen be Rs. y. According to the statement, we have 3x + 2y = 80 and 4x + 3y = 110 (ii) (b) x = 20, y = 10 3x + 2y = 3(20) + 2(10) = 60 + 20 = 80 4x + 3y = 4(20) + 3(10) = 80 + 30 = 110 (b) x = 20, y = 10 (iii) (b) Rs. 10 Cost of 1 pen = Rs. 10 (b) Rs. 10 (iv) (d) Rs. 420 Total cost = Rs. 15 x 20 + Rs. 12 x 10 = 300 + 120 = Rs. 420 (v) (a) Deepak Ram said that price of each notebook could be Rs. 25. Ajay felt that Rs. 2.50 for one pen was too little. It should be at least Rs. 16 Deepak guess the cost of one pen is Rs. 10 and Lohith guess the cost of one notebook is Rs. 30 Therefore, estimation of Deepak is correct

(a) (iii) x2 + x = 1 (b) (ii) x + y = 200 Here, x represents Sita's contribution and y represents Gita's contribution. (c) (iii) Rs. 124 If x = 76 then 76 + y = 200 y = 200 - 76 y = 124 (d) (ii) Rs. 100, Rs. 100 If x = y then x + x = 200 2x = 200 x = 200/2 = 100 (e) (iii) 1.x + 0.y + 5 = 0 Since, x = -5 ⇒ x + 5 = 0 Thus, standard form of x = -5 is 1.x + 0.y + 5 = 0.

(i) (d) 5x + 2y = 120 Here, the cost of one notebook be Rs. x and that of pen be Rs. y. According to the statement, we have 5x + 2y = 120 (ii) (b) x = 20, y = 10 5x + 2y = 5(10) + 2(20) = 50 + 40 = 90 ≠ 120 5x + 2y = 5(20) + 2(10) = 100 + 20 = 120 5x + 2y = 5(15) + 2(15) = 75 + 30 = 105 ≠ 120 (c) (ii) Rs. 95 5x + 2y = 5(15) + 2(10) = 75 + 20 = 95 (d) (b) Rs. 10 Here, x = 2y 5(2y) + 2y = 10y + 2y = 12y = 120 ⇒ y = 10 (e) (iv) 0.x + 1.y – 4 = 0 Since, y = 4 ⇒ y – 4 = 0 Thus, standard form of y = 4 is 0.x + 1.y – 4 = 0

(a) (iii) 2x + 3y = 60 Let the number of children be x and the number of adults be y then the linear equation in two variable for the given situation is  2x + 3y = 60. (b) (i) 10 2x + 3y =60 ⇒ 2(15) + 3y = 60 ⇒ 3y = 60 - 30 = 30  ⇒ y = 10 (c) (i) 12 2x + 3y = 60 ⇒ 2x + 3(12) = 60 ⇒ 2x 60 - 36 = 24 ⇒ x = 12 (d) (iii) 15 On putting x = 5 and y = 0 in the equation 3x + 5y = b, we have  3 x 5 + 5 x 0 = b ⇒ 15 + 0 = b ⇒ b = 15 (e) (ii) 1.x - 1.y + 5 = 0 y - x = 5 ⇒ y = x + 5 ⇒ x - y + 5 = 0 ⇒ 1.x - 1.y + 5 = 0

(a) (iii) x + y = 7 (b) (iv) 2 x + y = 7 ⇒ 5 + y = 7 ⇒ y = 7 - 5 = 2 (c) (i) (3, 4) (d) (iv) (0, 7) (e) (iii) -1 On putting x = 2 and y = -1 in the equation x + 3y - k = 0, we have 2 + 3(-1) - k = 0 ⇒ 2 - 3 = k ⇒ k = -1

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Course: Algebra 1   >   Unit 4

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10th Class Mathematics Pair of Linear Equations in Two Variables Question Bank

Done case based (mcqs) - pair of linear equations in two variables total questions - 40.

A) \[2x+y=19,\,\,x+y=13\] done clear

B) \[x+2y=19,\,\,x+y=18\] done clear

C) \[x+y=19,\,\,x+2y=13\] done clear

D) \[x+3y=19,2\,x+y=15\] done clear

question_answer 2) Find the perimeter of the outer boundary of the Layout

A) \[45\,m\] done clear

B) \[54m\] done clear

C) \[44m\] done clear

D) \[55m\] done clear

question_answer 3) Find the area of each bedroom and kitchen in the Layout

A) Area of each bedroom\[=\text{35 sq}.\text{ m}\] Area of kitchen \[=\text{3}0\text{ sq}.\text{ m}\] done clear

B) Area of each bedroom \[=\text{53 sq}.\text{ m}\] Area of kitchen \[=\text{3}0\text{ sq}.\text{ m}\] done clear

C) Area of each bedroom \[=\text{3}0\text{ sq}.\text{ m}\] Area of kitchen \[=\text{35 sq}.\text{ m}\] done clear

D) None of the above   done clear

question_answer 4) Find the area of living room in the layout

A) \[\text{75 sq}.\text{ m}\] done clear

B) \[\text{57 sq}.\text{ m}\]   done clear

C) \[\text{55 sq}.\text{ m}\] done clear

D) \[\text{77 sq}.\text{ m}\] done clear

question_answer 5) Find the cost of laying tiles in kitchen at the rate of Rs.50 per sq. m.

A) Rs.1250 done clear

B) Rs.1575 done clear

C) Rs.1700 done clear

D) Rs.1750 done clear

A) \[\text{8 km}/\text{h}\] done clear

B) \[\text{1}0\text{ km}/\text{h}\] done clear

C) \[\text{12 km}/\text{h}\] done clear

D) \[\text{14 km}/\text{h}\] done clear

question_answer 7) The speed of stream is:

A) \[\text{3 km}/\text{h}\] done clear

B) \[\text{4 km}/\text{h}\] done clear

C) \[\text{5 km}/\text{h}\] done clear

D) \[\text{6 km}/\text{h}\] done clear

question_answer 8) Which mathematical concept is used in above problem?

A) Pair of linear equations done clear

B) Cross-multiplication method done clear

C) Factorisation method done clear

D) None of the above done clear

question_answer 9) The direction in which the speed is maximum, is:

A) upstream done clear

B) downstream done clear

C) both have equal speed done clear

D) None of these   done clear

question_answer 10) The average speed of stream and boat in still water is:

A) \[\text{7 km}/\text{h}\] done clear

B) \[\text{10 km}/\text{h}\]   done clear

D) \[\text{5 km}/\text{h}\] done clear

A) \[\text{x}-\text{2y}=0\]and \[\text{3x}+\text{4y}=\text{2}0\] done clear

B) \[\text{x}+\text{2y}=0\]and \[\text{3x}-\text{4y}=\text{2}0\] done clear

C) \[\text{x}-\text{2y}=0\]and \[\text{4x}+\text{3y}=\text{2}0\] done clear

D) \[None\,\, of\,\, the\,\,above\] done clear

question_answer 12) Graphically, if the pair of equations intersect at one point, then the pair of equation is:      

A) Consistent done clear

B) inconsistent    done clear

C) consistent or inconsistent done clear

D) None of these  done clear

question_answer 13) The intersection point of two lines is:

A) \[(-4,-2)\] done clear

B) \[(4,3)\] done clear

C) \[(2,4)\] done clear

D) \[(4,2)\] done clear

question_answer 14) Intersection points of the line \[x-2y=0\]on x and y-axes are:

A) \[(2,0),\,\,(0,1)\] done clear

B) \[(1,0),\,\,(0,2)\] done clear

C) \[(0,0)\] done clear

D) \[None\,\, of\,\, these\] done clear

question_answer 15) Intersection points of the line \[\text{3x}+\text{4y}=\text{2}0\]on x and y-axes -are:

A) \[\left( \frac{20}{3},0 \right),\,(0,5)\] done clear

B) \[(2,0),\,\,(0,1)\] done clear

C) \[(5,0),\left( 0,\frac{20}{3} \right)\] done clear

A) Length = 6 m, breadth = 4 m done clear

B) length = 10 m, breadth = 6 m done clear

C) length =10 m, breadth =4 m done clear

D) length = 6 m, breadth = 2m done clear

question_answer 17) If the graphs of the equations in the given situation are plotted on the same graph paper, then:     

A) The lines will be parallel done clear

B) The lines will coincide done clear

C) The lines will intersect done clear

D) Can't say done clear

question_answer 18) The coordinates of the points where the two Lines, when plotted on a graph paper, intersect the x-axis are:

A) \[(4,0)\] and \[(8,0)\] done clear

B) \[(-4,0)\] and \[(8,0)\] done clear

C)  \[(4,0)\]and \[(-8,0)\] done clear

D) \[(-4,0)\] and \[(-8,0)\] done clear

question_answer 19) The value of k for which the system of equations \[\text{x+y}-\text{4=0}\]and \[\text{2x}+\text{ky}=\text{3}\] has no solution, is:

A) \[-2\] done clear

B) \[\ne 2\] done clear

C) 2 done clear

D) 3 done clear

question_answer 20) The equation of a line parallel to the line whose equation is given by \[\text{3x}-\text{2y}=\text{8}\] can be:

A) \[\text{3x}+\text{2y}=8\] done clear

B) \[3x-2y=8\] done clear

C) \[6x+4y=16\] done clear

D) \[15x-10y=45\] done clear

A) \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\] done clear

B) \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\] done clear

C) \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\] done clear

question_answer 22) If pair of lines are parallel, then pair of linear equations is:

A) inconsistent  done clear

B) consistent done clear

D) None of these done clear

question_answer 23) Check weather the two paths will cross each other or not.

A) yes done clear

B) no done clear

C) do not say done clear

question_answer 24) How many point(s) lie on the Line \[\text{x}-\text{3y}=\text{2}\]?

A) one done clear

B) two done clear

C) three done clear

D) infinitely done clear

question_answer 25) If the line \[\text{2x}+\text{6y}=\text{5}\] intersect the X-axis, then find its coordinate.

A) \[(-2.5,0)\] done clear

B) \[(2.5,0)\] done clear

C) \[(0,2.5)\] done clear

D) \[(0,-2.5)\] done clear

A) \[-\text{4x}-\text{3y}=-\text{4}\]and \[\text{4x}+\text{3y}=\text{2}0\] done clear

B) \[\text{4x}-\text{3y}=\text{4}\]and \[\text{4x}+\text{3y}=\text{2}0\] done clear

C) \[\text{4x}-\text{3y}=-\text{4}\]and \[\text{4x}+\text{3y}=\text{2}0\] done clear

D) \[-\text{4x}+\text{3y}=-\text{4}\]and \[\text{4x}+\text{3y}=\text{2}0\] done clear

question_answer 27) The cost of one ring game and one balloon game is:

A) Rs.2and Rs.4 done clear

B) Rs.4 and Rs.2 done clear

C) Rs.8 and Rs.2 done clear

D) Rs.6 and Rs.3 done clear

question_answer 28) The points where the line represented by the equation \[\text{4x}-\text{3y}=-\text{ 4}\] intersects the x-axis and y-axis, respectively are given by:

A) \[(1,0),\,\,\left( 0,\frac{4}{3} \right)\] done clear

B) \[(1,0),\,\,\left( 0,\frac{4}{3} \right)\] done clear

C) \[(-1,0),\,\,\left( 0,-\frac{4}{3} \right)\] done clear

D) \[(1,0),\,\,\left( 0,-\frac{4}{3} \right)\] done clear

question_answer 29) The area of the triangle formed by the two lines and the x-axis is:

A) 4 sq. units done clear

B) 6 sq. units done clear

C) 8 sq. units done clear

D) 12 sq. units done clear

question_answer 30) The value of k for which the pair of linear equations \[-x+y=-1,\] \[x+ky=5\] will be inconsistent, is:        

A) \[k=1\] done clear

B) \[k=-1\] done clear

C) \[k\ne -1\] done clear

D) \[k\ne 1\] done clear

A) \[\text{x}+\text{ y}=\text{2}000\]and \[x+2y=2800\] done clear

B) \[\text{x}+\text{2y}=\text{2}000\]and \[\text{x}+\text{y}=\text{28}00\] done clear

C) \[\text{2x}+\text{y}=\text{2}000\]and \[\text{x}+\text{y}=\text{28}00\] done clear

D) \[\text{x}+y=\text{2}000\]and \[\text{2x}+\text{y}=\text{28}00\] done clear

question_answer 32) Find the number of children and adults who bought tickets on that particular day are:

A) 1100 and 900 done clear

B) 1200 and 800 done clear

C) 1300 and 700 done clear

D) 1500 and 300 done clear

question_answer 33) Find the value(s) of k for which the pair of linear equations given by \[\text{2x}+\text{5y}=\text{2};\] \[(k+2)x+(2k+1)y=2(k-1)\] will have infinitely many solutions.

A) \[6\] done clear

B) \[-4\] done clear

C) \[4\] done clear

D) \[-6\]   done clear

question_answer 34) Find the points where the lines represented by the equations \[\text{2x}-\text{5y}+\text{4}=0\]and \[\text{x}+\text{2y}-\text{5}=0\]respectively intersect the x-axis.

A) \[(-3,0)\] and \[(4,0)\] done clear

B) \[(-2,0)\] and \[(5,0)\] done clear

C) \[(-1,0)\] and \[(6,0)\] done clear

D) \[(2,0)\] and \[(4,0)\] done clear

question_answer 35) Write the number of solutions of the system of linear equations \[\text{2x}-\text{3y}+\text{4}=0\] and \[\text{x}+\text{2y}-\text{5=0}\].

A) 1 done clear

B) 2 done clear

C) 0 done clear

D) infinite done clear

A) 24 done clear

B) 96 done clear

C) 70 done clear

D) 100 done clear

question_answer 37) How many questions did he guess?

B) 70       done clear

C) 100 done clear

D) 96 done clear

question_answer 38) If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?

A) 96 done clear

B) 24 done clear

D) 70 done clear

question_answer 39) If answer to all questions be attempted by guessing were wrong, then how many questions answered correctly to score 90 marks?

B) 70 done clear

C) 96 done clear

question_answer 40) If answer to all questions he attempter by guessing were wrong and answered \[\text{5}0%\]correctly, then how many marks he got?

A) 45 done clear

B) 55       done clear

C) 65 done clear

D) 40 done clear

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Case Based (MCQs) - Pair of Linear Equations in Two Variables

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• CBSE Maths Important Questions
• Class 9 Maths
• Chapter 4: Linear Equations Two Variables

Important Questions CBSE Class 9 Maths Chapter 4-Linear Equation in Two Variables

Important Questions of CBSE Class 9 Maths Chapter 4 -Linear equations in two variables with solutions are available for the students who are preparing 9th final exam. These problems are solved by our experts, as per NCERT book formulated by CBSE board. It covers all the questions according to the syllabus, which is important as per the exam point of view.

Students can reach us at BYJU’S to get important questions for all chapter CBSE class 9 Maths . Practising these extra questions will also help them to revise all the concepts and get good marks in the examination.

Also Check:

• Important 2 Marks Questions for CBSE 9th Maths
• Important 3 Marks Questions for CBSE 9th Maths
• Important 4 Marks Questions for CBSE 9th Maths

Important Questions & Solutions For Class 9 Maths Chapter 4 (Linear Equation in Two Variables)

Q.1: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) x – y/5 – 10 = 0

(ii) -2x+3y = 6

(iii) y – 2 = 0

(i) The equation x-y/5-10 = 0 can be written as:

(1)x + (-1/5) y + (-10) = 0

Now compare the above equation with ax + by + c = 0

Thus, we get;

(ii) –2x + 3y = 6

Re-arranging the given equation, we get,

–2x + 3y – 6 = 0

The equation –2x + 3y – 6 = 0 can be written as,

(–2)x + 3y +(– 6) = 0

Now comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0

We get, a = –2

(iii) y – 2 = 0

The equation y – 2 = 0 can be written as,

0x + 1y + (–2) = 0

Now comparing 0x + 1y + (–2) = 0with ax + by + c = 0

We get, a = 0

Q.2. Write four solutions for each of the following equations:

(i) 2x + y = 7

To find the four solutions of 2x + y = 7 we substitute different values for x and y

(2×0)+y = 7

(2×1)+y = 7

The solutions are (0, 7), (1,5), (3,1), (2,3)

(ii) πx + y = 9

To find the four solutions of πx + y = 9 we substitute different values for x and y

(π × 0)+y = 9

(π×1)+y = 9

(π(-1))+y = 9

The solutions are (0,9), (1,9-π),(9/π,0),(-1,9+π)

Q.3: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

The given equation is

2x + 3y = k

According to the question, x = 2 and y = 1.

Now, Substituting the values of x and y in the equation 2x + 3y = k,

⇒(2 x 2)+ (3 × 1) = k

The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.

Q.4: Draw the graph of each of the following linear equations in two variables:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values for which x and y satisfies the given equation.

Substituting the values for x,

When x = 0,

When x = 1,

The points to be plotted are (0, 0) and (1, 3)

(ii) 3 = 2x + y

⇒ 3 = 2(0) + y

⇒ 3 = 0 + y

⇒ 3 = 2(1) + y

⇒ 3 = 2 + y

⇒ y = 3 – 2

The points to be plotted are (0, 3) and (1, 1)

Q.5: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

3y = ax + 7

According to the question, x = 3 and y = 4

Now, Substituting the values of x and y in the equation 3y = ax + 7,

(3×4) = (ax3) + 7

⇒ 12 = 3a+7

⇒ 3a = 12–7

The value of a, if the point (3, 4) lies on the graph of the equation 3y = ax + 7 is 5/3.

Q.6: Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.

We have the equation,

For A (1, 2),

Substituting (x,y) = (1, 2),

2 = 9(1) – 7

For B (–1, –16),

Substituting (x,y) = (–1, –16),

–16 = 9(–1) – 7

-16 = – 9 – 7

For C (0, –7),

Substituting (x,y) = (0, –7),

– 7 = 9(0) – 7

Hence, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.

Thus, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7

Therefore, the points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7.

Q.7: Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?

Solution: Given equation,

3x + 4y = 6.

We need at least 2 points on the graph to draw the graph of this equation,

Thus, the points the graph cuts

Since the point is on the x-axis, we have y = 0.

Substituting y = 0 in the equation, 3x + 4y = 6,

3x + 4×0 = 6

Hence, the point at which the graph cuts x-axis = (2, 0).

(ii) y-axis

Since the point is on the y-axis, we have, x = 0.

Substituting x = 0 in the equation, 3x + 4y = 6,

3×0 + 4y = 6

Hence, the point at which the graph cuts y-axis = (0, 1.5).

Plotting the points (0, 1.5) and (2, 0) on the graph.

​Extra Questions For Class 9 Maths Chapter 4

• The taxi fare in a city is as follows: For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
• in one variable
• in two variables
• Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.

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