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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 4 - Linear Equations in Two Variables

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 4 - Linear Equations in Two Variables.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 4 - Linear Equations in Two Variables Page/Excercise MCQ

Solution 1

Correct option: (b)

The equation of the x-axis is y = 0.

Solution 2

Correct option: (a)

The equation of the y-axis is x = 0. 

Solution 3

  

Solution 4

  

Solution 5

  

Solution 6

Correct option: (a)

The equation 2x + 5y = 7 has a unique solution, if x and y are natural numbers.

If we take x = 1 and y = 1, the given equation is satisfied. 

Solution 7

Correct option: (c)

The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin. 

Solution 8

Correct option: (d)

The graph of x = 4 is a line parallel to the y-axis at a distance of 4 units from the origin. 

Solution 9

Correct option: (c)

The graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis. 

Solution 10

Correct option: (c)

The graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis. 

Solution 11

Correct option: (c)

When a graph meets the y-axis, the x coordinate is zero.

Thus, substituting x = 0 in the given equation, we get

2(0) + 3y = 6

3y = 6

y = 2

Hence, the required point is (0, 2).

Solution 12

Correct option: (c)

When a graph meets the x-axis, the y coordinate is zero.

Thus, substituting y = 0 in the given equation, we get

2x + 5(0) = 10

2x = 10

x = 5

Hence, the required point is (5, 0). 

Solution 13

  

  

Solution 14

Correct option: (c)

Since, the y coordinate is 3, the graph of the line y = 3 passes through the point (2, 3).

Solution 15

   

Solution 16

  

Solution 17

  

Solution 18

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2 and y = 3. 

Solution 19

  

Solution 20

Correct option: (d)

Since, (2, 0) is a solution of the linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given equation, we have

2(2) + 3(0) = k

4 + 0 = k

k = 4 

Solution 21

  

Solution 22

  

Solution 23

Correct option: (c)

Substituting x = 5 and y = 2 in L.H.S. of equation x + y = 7, we get

L.H.S. = 5 + 2 = 7 = R.H.S.

Hence, x = 5 and y = 2 is a solution of the linear equation x + y = 7. 

Solution 24

Correct option: (b)

Since the point (3, 4) lies on the graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we get

3(4) = a(3) + 7

12 = 3a + 7

3a = 5

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 4 - Linear Equations in Two Variables Page/Excercise 4A

Solution 1(i)

We have,

3x + 5y = 7.5

3x + 5y - 7.5 = 0

6x + 10y - 15 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = 10 and c = -15 

Solution 1(ii)

  

On comparing this equation with ax + by + c = 0, we obtain

a = 10, b = -1 and c = 30 

Solution 1(iii)

We have,

3y - 2x = 6

-2x + 3y - 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = -2, b = 3 and c = -6 

Solution 1(iv)

We have,

4x = 5y

4x - 5y = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 4, b = -5 and c = 0 

Solution 1(v)

  

 6x - 5y = 30

 6x - 5y - 30 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 6, b = -5 and c = -30 

Solution 1(vi)

  

On comparing this equation with ax + by + c = 0, we obtain

a =  , b =   and c = -5 

Solution 2(i)

We have,

x = 6

x - 6 = 0

1x + 0y - 6 = 0

x + 0y - 6 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 0 and c = -6 

Solution 2(ii)

We have,

3x - y = x - 1

3x - x - y + 1 = 0

2x - y + 1 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = -1 and c = 1 

Solution 2(iii)

We have,

2x + 9 = 0

2x + 0y + 9 = 0

On comparing this equation with ax + by + c = 0, we obtain

a = 2, b = 0 and c = 9 

Solution 2(iv)

We have,

4y = 7

0x + 4y - 7 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 0, b = 4 and c = -7 

Solution 2(v)

We have,

x + y = 4

x + y - 4 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 1, b = 1 and c = -4 

Solution 2(vi)

We have,

  

3x - 8y - 1 = 0 

On comparing this equation with ax + by + c = 0, we obtain

a = 3, b = -8 and c = -1 

Solution 3(i)

Given equation is 5x - 4y = 20

Substituting x = 4 and y = 0 in L.H.S. of given equation, we get

L.H.S. = 5x - 4y

= 5(4) - 4(0)

= 20 - 0

= 20

= R.H.S.

Hence, (4, 0) is the solution of the given equation.

Solution 3(ii)

Given equation is 5x - 4y = 20

Substituting x = 0 and y = 5 in L.H.S. of given equation, we get

L.H.S. = 5x - 4y

= 5(0) - 4(5)

= 0 - 20

= -20

R.H.S.

Hence, (0, 5) is not the solution of the given equation. 

Solution 3(iii)

Given equation is 5x - 4y = 20

Substituting x = -2 and y =   in L.H.S. of given equation, we get

L.H.S. = 5x - 4y

= 5(-2) - 4

= -10 - 10

= -20

R.H.S.

Hence,   is not the solution of the given equation. 

Solution 3(iv)

Given equation is 5x - 4y = 20

Substituting x = 0 and y = -5 in L.H.S. of given equation, we get

L.H.S. = 5x - 4y

= 5(0) - 4(-5)

= 0 + 20

= 20

= R.H.S.

Hence, (0, -5) is the solution of the given equation. 

Solution 3(v)

Given equation is 5x - 4y = 20

Substituting x = 2 and y =   in L.H.S. of given equation, we get

L.H.S. = 5x - 4y

= 5(2) - 4

= 10 + 10

= 20

= R.H.S.

Hence,   is the solution of the given equation. 

Solution 4(a)

Given equation is 2x - 3y = 6

Substituting x = 0 in the given equation, we get

2(0) - 3y = 6

0 - 3y = 6

3y = -6

y = -2

So, (0, -2) is the solution of the given equation.

 

Substituting y = 0 in the given equation, we get

2x - 3(0) = 6

2x - 0 = 6

2x = 6

x = 3

So, (3, 0) is the solution of the given equation.

 

Substituting x = 6 in the given equation, we get

2(6) - 3y = 6

12 - 3y = 6

3y = 6

y = 2

So, (6, 2) is the solution of the given equation.

 

Substituting y = 4 in the given equation, we get

2x - 3(4) = 6

2x - 12 = 6

2x = 18

x = 9

So, (9, 4) is the solution of the given equation.

 

Substituting x = -3 in the given equation, we get

2(-3) - 3y = 6

-6 - 3y = 6

3y = -12

y = -4

So, (-3, -4) is the solution of the given equation.

Solution 4(b)

Given equation is   

Substituting x = 0 in (i), we get

4(0) + 3y = 30

3y = 30

y = 10

So, (0, 10) is the solution of the given equation.

 

Substituting x = 3 in (i), we get

4(3) + 3y = 30

12 + 3y = 30

3y = 18

y = 6

So, (3, 6) is the solution of the given equation.

 

Substituting x = -3 in (i), we get

4(-3) + 3y = 30

-12 + 3y = 30

3y = 42

y = 14

So, (-3, 14) is the solution of the given equation.

 

Substituting y = 2 in (i), we get

4x + 3(2) = 30

4x + 6 = 30

4x = 24

x = 6

So, (6, 2) is the solution of the given equation.

 

Substituting y = -2 in (i), we get

4x + 3(-2) = 30

4x - 6 = 30

4x = 36

x = 9

So, (9, -2) is the solution of the given equation.

Solution 4(c)

Given equation is 3y = 4x

Substituting x = 3 in the given equation, we get

3y = 4(3)

3y = 12

y = 4

So, (3, 4) is the solution of the given equation.

 

Substituting x = -3 in the given equation, we get

3y = 4(-3)

3y = -12

y = -4

So, (-3, -4) is the solution of the given equation.

 

Substituting x = 9 in the given equation, we get

3y = 4(9)

3y = 36

y = 12

So, (9, 12) is the solution of the given equation.

 

Substituting y = 8 in the given equation, we get

3(8) = 4x

4x = 24

x = 6

So, (6, 8) is the solution of the given equation.

 

Substituting y = -8 in the given equation, we get

3(-8) = 4x

4x = -24

x = -6

So, (-6, -8) is the solution of the given equation.

Solution 5

Since x = 3 and y = 4 is a solution of the equation 5x - 3y = k, substituting x = 3 and y = 4 in equation 5x - 3y = k, we get

5(3) - 3(4) = k

15 - 12 = k

k = 3 

Solution 6

Since x = 3k + 2 and y = 2k - 1 is a solution of the equation 4x - 3y + 1 = 0, substituting these values in equation, we get

4(3k + 2) - 3(2k - 1) + 1 = 0

12k + 8 - 6k + 3 + 1 = 0

6k + 12 = 0

6k = -12

k = -2 

Solution 7

Let the cost of one pencil be Rs. x and that of one ballpoint be Rs. y.

Then,

Cost of 5 pencils = Rs. 5x

Cost of 2 ballpoints = Rs. 2y

According to given statement, we have

5x = 2y

5x - 2y = 0

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 4 - Linear Equations in Two Variables Page/Excercise 4B

Solution 1(i)

x = 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to its right.

  

Solution 1(ii)

x + 4 = 0

x = -4, which is a line parallel to the Y-axis, at a distance of 4 units from it, to its left.

  

Solution 1(iii)

y = 3 is a line parallel to the X-axis, at a distance of 3 units from it, above the X-axis.

  

Solution 1(iv)

y = -3 is a line parallel to the X-axis, at a distance of 3 units from it, below the X-axis.

  

Solution 1(v)

x = -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to its left.

  

Solution 1(vi)

x = 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to its right.

  

Solution 1(vii)

y + 5 = 0

y = -5, which is a line parallel to the X-axis, at a distance of 5 units from it, below the X-axis.

  

Solution 1(viii)

y = 4 is a line parallel to the X-axis, at a distance of 4 units from it, above the X-axis.

  

Solution 2(ii)

The given equation is y = 3x.

Putting x = 1, y = 3 1 = 3

Putting x = 2, y = 3 2 = 6

Thus, we have the following table:

x

1

2

y

3

6

Plot points (1,3) and (2,6) on a graph paper and join them to get the required graph.

Take a point P on the left of y-axis such that the distance of point P from the y-axis is 2 units.

Draw PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to x-axis meeting y-axis at N.

So, y = ON = -6.

Solution 2(i)

y = 3x

When x = 1, then y = 3(1) = 3

When x = -1, then y = 3(-1) = -3

Thus, we have the following table:

x

1

-1

y

3

-3

 

Now, plot the points A(1, 3) and B(-1, -3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of y = 3x.

  

Reading the graph

Given: x = 2. Take a point M on the X-axis such that OM = 2.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 6

Thus, when x = 2, then y = 6.

Solution 3(i)

The given equation is,

x + 2y - 3 = 0

x = 3 - 2y

Putting y = 1,x = 3 - (2 1) = 1

Putting y = 0,x = 3 - (2 0) = 3

Thus, we have the following table:

x

1

3

y

1

0

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 - 2y) at P.

Through P, draw PM parallel to x-axis cutting y-axis at M.

So, y = OM = -1.

Solution 3(ii)

x + 2y - 3 = 0

2y = 3 - x

When x = -1, then   

When x = 1, then

Thus, we have the following table:

x

-1

1

y

2

1

 

Now, plot the points A(-1, 2) and B(1, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + 2y - 3 = 0.

  

Reading the graph

Given: x = -5. Take a point M on the X-axis such that OM = -5.

Draw MP parallel to the Y-axis, cutting the line AB at P.

Clearly, PM = 4

Thus, when x = -5, then y = 4. 

Solution 4

The given equation is, 2x - 3y = 5

Now, if x = 4, then

And, if x = -2, then

Thus, we have the following table:

x

4

-2

y

1

-3

Plot points (4,1) and (-2,-3) on a graph paper and join them to get the required graph.

(i) When x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to its right cutting the line at Q and through Q draw a line parallel to x-axis cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

(ii) When y = 3, draw a line parallel to x-axis at a distance of 3 units from x-axis and above it, cutting the line at point P. Through P, draw a line parallel to y-axis meeting x-axis at a point which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.

Solution 5

The given equation is 2x + y = 6

y = 6 - 2x

Now, if x = 1, then y = 6 - 2 1 = 4

And, if x = 2, then y = 6 - 2 2 = 2

Thus, we have the following table:

x

1

2

y

4

2

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).

Solution 6

The given equation is 3x + 2y = 6

2y = 6 - 3x

Now, if x = 2, then

And, if x = 4, then

Thus, we have the following table:

x

2

4

y

0

-3

Plot points (2, 0) and (4,-3) on a graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6 cuts the y-axis at a point P which is 3 units above the x-axis.

So, co-ordinates of P are (0,3).

Solution 7

Graph of the equation 3x - 2y = 4

2y = 3x - 4

When x = 2, then   

When x = -2, then

Thus, we have the following table:

x

2

-2

y

1

-5

 

Now, plot the points A(2, 1) and B(-2, -5) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 3x - 2y = 4.

 

Graph of the equation x + y - 3 = 0

y = 3 - x

When x = 1, then y = 3 - 1 = 2 

When x = -1, then y = 3 - (-1) = 4

Thus, we have the following table:

x

1

-1

y

2

4

 

Now, plot the points C(1, 2) and D(-1, 4) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x + y - 3 = 0.

  

The two graph lines intersect at point A(2, 1). 

Solution 8(i)

4x + 3y = 24

3y = 24 - 4x

When x = 0, then   

When x = 3, then

Thus, we have the following table:

x

0

3

y

8

4

 

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

  

Reading the graph

The graph of line 4x + 3y = 24 intersects the X-axis at point C(6, 0) and the Y-axis at point A(0, 8). 

Solution 8(ii)

4x + 3y = 24

3y = 24 - 4x

When x = 0, then   

When x = 3, then

Thus, we have the following table:

x

0

3

y

8

4

 

Now, plot the points A(0, 8) and B(3, 4) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 4x + 3y = 24.

  

Reading the graph

Required area = Area of ΔAOC

Solution 9

Graph of the equation 2x + y = 6

y = 6 - 2x

When x = 1, then y = 6 - 2(1) = 6 - 2 = 4 

When x = 2, then y = 6 - 2(2) = 6 - 4 = 2

Thus, we have the following table:

x

1

2

y

4

2

 

Now, plot the points A(1, 4) and B(2, 2) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of 2x + y = 6.

 

Graph of the equation 2x - y + 2 = 0

y = 2x + 2

When x = -1, then y = 2(-1) + 2 = -2 + 2 = 0 

When x = 2, then y = 2(2) + 2 = 4 + 2 = 6

Thus, we have the following table:

x

-1

2

y

0

6

 

Now, plot the points C(-1, 0) and D(2, 6) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x - y + 2 = 0.

  

The two graph lines intersect at point A(1, 4).

The area enclosed by the lines and X-axis is shown in the graph.

Draw AM perpendicular from A on X-axis.

PM = y-coordinate of point A(1, 4) = 4

And, CP = 4

Area of shaded region = Area of ΔACP

  

Solution 10

Graph of the equation x - y = 1

y = x - 1

When x = 1, then y = 1 - 1 = 0 

When x = 2, then y = 2 - 1 = 1

Thus, we have the following table:

x

1

2

y

0

1

 

Now, plot the points A(1, 0) and B(2, 1) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x - y = 1.

 

Graph of the equation 2x + y = 8

y = 8 - 2x

When x = 2, then y = 8 - 2(2) = 8 - 4 = 4 

When x = 3, then y = 8 - 2(3) = 8 - 6 = 2 

Thus, we have the following table:

x

2

3

y

4

2

 

Now, plot the points C(2, 4) and D(3, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of 2x + y = 8.

  

The two graph lines intersect at point D(3, 2).

The area enclosed by the lines and Y-axis is shown in the graph.

Draw DM perpendicular from D on Y-axis.

DM = x-coordinate of point D(3, 2) = 3

And, EF = 9

Area of shaded region = Area of ΔDEF

Solution 11

Graph of the equation x + y = 6

y = 6 - x

When x = 2, then y = 6 - 2 = 4 

When x = 3, then y = 6 - 3 = 3

Thus, we have the following table:

x

2

3

y

4

3

 

Now, plot the points A(2, 4) and B(3, 3) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 6.

 

Graph of the equation x - y = 2

y = x - 2

When x = 3, then y = 3 - 2 = 1 

When x = 4, then y = 4 - 2 = 2 

Thus, we have the following table:

x

3

4

y

1

2

 

Now, plot the points C(3, 1) and D(4, 2) on a graph paper.

Join CD and extend it in both the directions.

Then, the line CD is the required graph of x - y = 2.

  

The two graph lines intersect at point D(4, 2).

Solution 12

Let the amount contributed by students A and B be Rs. x and Rs. y respectively.

Total contribution = 100

x + y = 100

y = 100 - x

When x = 25, then y = 100 - 25 = 75

When x = 50, then y = 100 - 50 = 50

Thus, we have the following table:

x

25

50

y

75

50

 

Now, plot the points A(25, 75) and B(50, 50) on a graph paper.

Join AB and extend it in both the directions.

Then, the line AB is the required graph of x + y = 100.

  

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 4 - Linear Equations in Two Variables for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 4 - Linear Equations in Two Variables now.

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CBSE IX - Mathematics

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