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RD Sharma Solution for Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 7 - 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 RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics 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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RD Sharma Solution for Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables Page/Excercise 7.1

Solution 1


Solution 2

Solution 3

RD Sharma Solution for Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables Page/Excercise 7.2

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7(i)

Solution 7(ii)

Solution 7(iii)

RD Sharma Solution for Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables Page/Excercise 7.3

Solution 1(i)


Solution 1(ii)


Solution 1(iii)


Solution 1(iv)


Solution 1(v)


Solution 1(vi)


Solution 1(vii)


Solution 1(viii)


Solution 2

Solution 3

Solution 4

The given points on the graph:



It is dear from the graph, the straight line passing through these points also passes through the point (1,4).

Solution 5



Solution 6



Solution 7

Solution 8





Solution 9(i)

Solution 9(ii)

Solution 9(iii)

Solution 9(iv)

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

We have,

y = |X|                               ...(i)

Putting x = 0, we get y = 0

Putting x = 2, we get y = 2

Putting x = -2, we get y = 2

Thus, we have the following table for the points on graph of |x|.

x 0 2 -2
y 0 2 2



The graph of the equation y = |x|:

Solution 16

We have,

y = |x| + 2                                                  ...(i)

Putting x = 0, we get y = 2

Putting x = 1, we get y = 3

Putting x = -1, we get y = 3

Thus, we have the following table for the points on graph of |x| + 2:

x 0 1 -1
y 2 3 3



The graph of the equation y = |x| + 2:

Solution 17




Solution 18



Solution 19


Solution 20


Solution 21



RD Sharma Solution for Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables Page/Excercise 7.4

Solution 1(i)


Solution 1(ii)



y + 3 = 0

y = -3

Point A represents -3 on number line.

On Cartesian plane, equation represents all points on x axis for which y = -3

Solution 1(iii)



y = 3

Point A represents 3 on number line.

On Cartesian plane, equation represents all points on x axis for which y = 3

Solution 1(iv)


Solution 1(v)


Solution 2(i)



Solution 2(ii)




Solution 3(i)



Solution 3(ii)



On Cartesian plane, equation represents all points on y axis for which x = -5

Solution 4

(i) The equation of the line that is parallel to x-axis and passing through the point (0,3) is y = 3

(ii) The equation of the line that is parallel to x-axis and passing through the point (0,-4) is y = -4

(iii) The equation of the line that is parallel to x-axis and passing through the point (2,-5) is y = -5

(iv) The equation of the line that is parallel to x-axis and passing through the point (3, 4) is y = 4

Solution 5

RD Sharma Solution for Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables Page/Excercise 7.33

Solution 1

y = ax + 3

If (4, 19) is its solution, then it must satisfy the equation.

Thus, we have

19 = a × 4 + 3

i.e. 4a = 16

i.e. a = 4

Hence, correct option is (b).

Solution 2

3x + y = 10

If (a, 4) lies on its graph, then it must satisfy the equation.

Thus, we have

3(a) + 4 = 10

i.e. 3a = 6

i.e. a = 2

Hence, correct option is (c).

Solution 3

On x-axis, the y-co-ordinate is always 0.

So, 2x - y = 4 will cut the x-axis where y = 0

i.e. 2x = 4

i.e. x = 2

Thus, 2x - y = 4 will cut the x-axis at (2, 0).

Hence, correct option is (a).

Solution 4

From Point (2, -3) there are infinitely many lines passing in every-direction.

So (2, -3) is satisfied with infinite linear equations.

Hence, correct option is (d).

Solution 5

Given equation is x – 2 = 0.

If this is treated as an equation in one variable x only, then it has the unique solution x = 2, which is a point on the number line. 

However, when treated as an equation in two variables, it can be expressed as x - 2 = 0.

So as, an equation in two variables, x – 2 = 0 is represented by a single line parallel to y-axis at a distance of 2 units.

Hence, correct option is (a).

Solution 6

Substituting x = 2 and y = -1 in the following equations:

L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 = R.H.S.

L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 ≠ 4 ≠ R.H.S.

L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 0 ≠ R.H.S.

L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 5 ≠ R.H.S.

Hence, correct option is (a).

Solution 7

If (2k - 1, k) is solution of equation 10x - 9y = 12, then it must satisfy this equation.

Thus, we have

10(2k - 1) - 9k = 12

20k - 10 - 9k = 12

11k = 22

k = 2

Hence, correct option is (b).

Solution 8

The distance between the graph of the equations x = -3 and x = 2

= 2 - (-3)

= 2 + 3

= 5

Hence, correct option is (d).

Solution 9

The distance between given two graphs

= 3 - (-1)

= 3 + 1

= 4

Hence, correct option is (b).

Solution 10

begin mathsize 12px style 4 straight x space plus space 3 straight y space equals space 12
At space straight x space equals space 0 comma space 3 straight y space equals space 12 rightwards double arrow straight y space equals space 4 space units
At space straight y space equals space 0 comma space 4 straight x space equals space 12 rightwards double arrow straight x space equals space 3 space units
The space triangle space formed space is space triangle AOB comma space where
OB space equals space 4 space units
OA space equals space 3 space units
Hypotenuse space equals space AB space equals space square root of OB squared space plus space OA squared end root space equals space square root of 16 space plus space 9 end root equals 5 space units
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 7 - Linear Equations in Two Variables for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 7 - Linear Equations in Two Variables.

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