RS Aggarwal Solutions Class 10 Linear Equations In Two Variables

RS Aggarwal Solutions Class 10 Chapter 3

All these RS Aggarwal class 10 solutions Chapter 3 Linear equations in two variables are solved by Byju's top ranked professors as per CBSE guidelines.

Linear Equation in Two Variables Exercise 3.1
Linear Equation in Two Variables Exercise 3.2
Linear Equation in Two Variables Exercise 3.3
Linear Equation in Two Variables Exercise 3.4
Linear Equation in Two Variables Exercise 3.5
Linear Equation in Two Variables Exercise 3.6
Linear Equation in Two Variables Exercise 3.7
Linear Equation in Two Variables Exercise 3.8
Linear Equation in Two Variables Exercise 3.9
Linear Equation in Two Variables Exercise 3.10
Linear Equation in Two Variables Exercise 3.11

Question 1: On a graph paper, draw a horizontal line X???OX and a vertical line YOY??? as the x-axis and the y-axis respectively.

2x + 3y = 2,??x – 2y = 8

Solution:

Given equations are 2x + 3y = 2 and x – 2y = 8

Graph of 2x + 3y = 2:

\( y = \frac{2(1 – x)}{3} \)

Putting x = 1, we get y = 0

Putting x = -2, we get y = 2

Putting x = 4, we get y = -2

Hence, the table is:

x 1 -2 4
y 0 2 -2

Plot the points A (1, 0), B (-2, 2) and C (4, -2) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x + 3y = 2.

Graph of x – 2y = 8:

\( y = \frac{x – 8}{2} \)

Putting x = 2, we get y = -3

Putting x = 4, we get y = -2

Putting x = 0, we get y = -4

Hence, the table is:

x 2 4 0
y -3 -2 -4

Now, on the same graph paper as above plot the points P (0, -4) and Q (2, -3). The point C (4, -2) has already been plotted. Join QC and extend it. Thus, the line PC is the graph of x – 2y = 8.

 

https://lh3.googleusercontent.com/cM08u_bTM44-sCE0syRfn8GzN1T9dJRE9DR56GQM2miO6tdI5NikKmTkDMr00lh40YUh8Ei0DUKnW531yU-m4DUSD-1OeLjExYkWXI_C0OrIq90IXNfSvFM4MDFJxDdWjqJL4FZR

 

The two graph lines intersect at C (4, -2). Therefore, x = 4, y = -2 is the solution of given system of equations.

 

Question 2: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

3x + 2y = 4 and 2x – 3y = 7

Solution:

Given equations are 3x + 2y = 4 and 2x – 3y = 7

Graph of 3x + 2y = 4:

\( y = \frac{4 – 3x}{2} \)

Putting x = 0, we get y = 2

Putting x = 2, we get y = -1

Putting x = -2, we get y = 5

Hence, the table is:

x 0 2 -2
y 2 -1 5

Plot the points A (0, 2), B (2, -1) and C (-2, 5) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 3x + 2y = 4.

Graph of 2x – 3y = 7:

\( y = \frac{x – 8}{2} \)

Putting x = 2, we get y = -1

Putting x = -1, we get y = -3

Putting x = 5, we get y = 1

Hence, the table is:

x 2 -1 5
y -1 -3 1

Now, on the same graph paper as above plot the points P (-1, -3) and Q (5, 1). The point B (2, -1) has already been plotted. Join QB and extend it. Thus, the line PQ is the graph of 2x – 3y = 7.

 

https://lh5.googleusercontent.com/bZEdupG_uRd3PY8RCGnZKvXcXpTFq3PZhJr2UdJ09Ht4yYZXz0Um3GKWZJo7rVxCLonkHE6MYzXfub_LEOJcWmfqbgCBddN5d7PJrcpFJtLffY77a9VyGqTBf11i9zZ4ZOqNrvrZ

 

The two graph lines intersect at B (2,-1). Therefore, x = 2, y = -1 is the solution of given system of equations.

 

Question 3: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

x – y + 1 = 0,

3x + 2y – 12 = 0

Solution:

Given equations are x – y + 1 = 0 and 3x + 2y – 12 = 0

Graph of x – y + 1 = 0:

x – y + 1 = 0 \( \Rightarrow \) y = x + 1 ??????..(1)

Putting x = 0, we get y = 1

Putting x = -1, we get y = 0

Putting x = 2, we get y = 3

Hence, the table is:

x 0 -1 2
y 1 0 3

Plot the points A (0, 1), B (-1, 0) and C (2, 3) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of x – y + 1 = 0.

Graph of 3x + 2y – 12 = 0:

\( y = \frac{12 – 3x}{2} \)

Putting x = 0, we get y = 6

Putting x = 2, we get y = 3

Putting x = 4, we get y = 0

Hence, the table is:

x 0 2 4
y 6 3 0

Now, on the same graph paper as above plot the points P (0, 6) and Q (4, 0). The point C (2, 3) has already been plotted. Join QB and extend it. Thus, the line PQ is the graph of 3x + 2y – 12 = 0.

 

https://lh3.googleusercontent.com/ULpP8_HD_q2rGBD0521Mt_YUrRgB2oDi_OHRZ1jfoav1e3M_wQ1WhVN4LcUwouZDFbgTdOhEMiuqN9GqYwbv7IB0KuAWmKzl6nLVMfzTqOvnrShQ7IkdHp8rtmv9Lg5p_AC2yCOT

 

The two graph lines intersect at (2, 3). Therefore, x = 2, y = 3 is the solution of given system of equations.

 

Question 4: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

2x + 3y = 4,??3x – y = -5

Solution:

Given equations are 2x + 3y = 4 and 3x – y = -5

Graph of 2x + 3y = 4:

2x + 3y = 4 \( \Rightarrow y = \frac{4 – 2x}{3} \)

Putting x = -1, we get y = 2

Putting x = 2, we get y = 0

Putting x = 5, we get y = -2

Hence, the table is:

x -1 2 5
y 2 0 -2

Plot the points A (-1, 2), B (2, 0) and C (5, -2) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line BC is the graph of 2x + 3y = 4.

Graph of 3x – y = -5:

y = 3x + 5

Putting x = -1, we get y = -2

Putting x = 0, we get y = 5

Putting x = -2, we get y = -1

Hence, the table is:

x -1 0 -2
y 2 5 -1

Now, on the same graph paper as above plot the points P(0, 5) and Q(-2, -1). The point A (-1, 2) has already been plotted. Join PA and amd QA to get line PQ. Thus, the line PQ is the graph of 3x – y = -5.

 

https://lh5.googleusercontent.com/d9qrqYfS9aoc10TnwimJLNQAXhux4JNtK15zuEmnhnL45dFeqfOg3pzQ6UtC9eXSkUjue7qQlH4eeMO13kcg7s1KNrZX7T07oWUTnpBfN1FZM4-e7M59V9WoovpxEHGmqlwk-woG

 

The two graph lines intersect at A (-1, 2). Therefore, x = -1, y = 2 is the solution of given system of equations.

 

Question 5: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

2x – 3y = 1,??3x – 4y = 1

Solution:

Given equations are 2x – 3y = 1 and 3x – 4y = 1

Graph of 2x – 3y = 1:

2x – 3y = 1 \( \Rightarrow y = \frac{2x – 1}{3} \)

Putting x = -1, we get y = -1

Putting x = 2, we get y = 1

Putting x = 5, we get y = 3

Hence, the table is:

x -1 2 5
y -1 1 3

Plot the points A (-1, -1), B (2, 1) and C (5, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2x – 3y = 1.

Graph of 3x – 4y = 1:

3x – 4y = 1 \( \Rightarrow y = \frac{3x – 1}{4} \)

Putting x = -1, we get y = -1

Putting x = 3, we get y = 2

Putting x = 2, we get y = -4

Hence, the table is:

x -1 3 -5
y -1 2 -4

Now, on the same graph paper as above plot the points P (3, 2) and Q(-5, -4). The point A (-1, -1) has already been plotted. Join PA and QA. Thus, the line PQ is the graph of 3x – 4y = 1.

 

https://lh3.googleusercontent.com/kFhDbEgER5IAP4neHsYAAsH5ouSDPr5JmQloYiar2TWT64TCJ1dUfzIFg9H_wotXr4zviihLl2YAc9a1hF7Fc43E52NSXepzUt0Zrpuiblsgs6r3wBoyzjMNpnNpNxBJ0uGstd1A

 

The two graph lines intersect at A (-1, -1). Therefore, x = -1, y = -1 is the solution of given system of equations.

 

Question 6: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

4x + 3y = 5,??2y – x = 7

Solution:

Given equations are 4x + 3y = 5 and 2y – x = 7

Graph of 4x + 3y = 5:

4x + 3y = 5 \( \Rightarrow y = \frac{5 – 4x}{3} \)

Putting x = -1, we get y = 3

Putting x = 2, we get y = -1

Putting x = 5, we get y = -5

Hence, the table is:

x -1 2 5
y 3 -1 -5

Plot the points A (-1, 3), B (2, -1) and C (5, -5) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x + 3y = 5.

Graph of 2y – x = 7:

2y – x = 7\( \Rightarrow y = \frac{7 + x}{2} \)

Putting x = -1, we get y = 3

Putting x = 3, we get y = 5

Putting x = -3, we get y = 2

Hence, the table is:

x -1 3 -3
y 3 5 2

Now, on the same graph paper as above plot the points P (3, 5) and Q (-3, 2). The point A (-1, 3) has already been plotted. Join PA and PQ. Thus, the line PQ is the graph of 2y – x = 7.

 

https://lh4.googleusercontent.com/hzQwQNwJYZCk2A1p8cfOiLKqLV5sLNYPj0u-U1AkQbPf4YSJshGVFsY9IpHBzE4L_h1yO2wmgZXCZBCsYPQzmQu5D7YpFBaA0zVcGEJ3p-6Vvg9bjTGbDVq5PYA2QNKUGm6p12lB

 

The two graph lines intersect at A(-1, -3). Therefore, x = -1, y = 3 is the solution of given system of equations.

 

Question 7: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

x + 2y + 2 = 0,??3x + 2y + 2 = 0

Solution:

Given equations are x + 2y + 2 = 0 and 3x + 2y + 2 = 0

Graph of x + 2y + 2 = 0:

x + 2y + 2 = 0 \( \Rightarrow y = \frac{-x -2}{2} \)

Putting x = -2, we get y = 0

Putting x = 0, we get y = -1

Putting x = 2, we get y = -2

Hence, the table is:

x -2 0 2
y 0 -1 -2

Plot the points A (-2, 0), B (0, -1) and C (2, -2) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of x + 2y + 2 = 0.

Graph of 3x + 2y + 2 = 0:

3x + 2y + 2 = 0 \( \Rightarrow y = \frac{-3x + 2}{2} \)

Putting x = 0, we get y = 1

Putting x = 2, we get y = -2

Putting x = 4, we get y = -5

Hence, the table is:

x 0 2 4
y 1 -2 -5

Now, on the same graph paper as above plot the points P (0, 1) and Q (4, -5). The point C (2, -2) has already been plotted. Join PC and QC. Thus, the line PQ is the graph of 3x + 2y + 2 = 0.

The two graph lines intersect at C(2, -2). Therefore, x = 2, y = -2 is the solution of given system of equations.

 

Question 8: On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.

2x + 3y = 8,??x – 2y + 3 = 0

Solution:

Given equations are 2x + 3y = 8 and x – 2y + 3 = 0

Graph of 2x + 3y = 8:

2x + 3y = 8 \( \Rightarrow y = \frac{8 – 2x}{3} \)

Putting x = 1, we get y = 2

Putting x = -5, we get y = 6

Putting x = 7, we get y = -2

Hence, the table is:

x 1 -5 7
y 2 6 -2

Plot the points A (1, 2), B (-5, 6) and C (7, -2) on the graph paper. Join AB and BC to get the graph line BC. Extend it both ways.

Thus, line AC is the graph of 2x + 3y = 8.

Graph of x – 2y + 3 = 0:

x – 2y + 3 = 0 \( \Rightarrow y = \frac{x + 3}{2} \)

Putting x = 1, we get y = 2

Putting x = 3, we get y = 3

Putting x = -3, we get y = 0

Hence, the table is:

x 1 3 -3
y 2 3 0

Now, on the same graph paper as above plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join PA and QA. Thus, the line PQ is the graph of x – 2y + 3 = 0.

 

https://lh4.googleusercontent.com/dFZ66sxeED3RfynET_9aXfMXKWJZJ3qow6KO9daIj7MCBa2a0hdZOA1otZLldT13Rmw6tSJfvQ7GrriqJwvywGU_zcVTC8YLWugflMMOZzAqM1didlujHrqTi8fVVD251MmRY3Yp

 

The two graph lines intersect at A (1, 2). Therefore, x = 1, y = 2 is the solution of given system of equations.

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