### RS Aggarwal Solutions Class 9 Chapter 8 Linear Equations in Two Variables

Here you can get solutions of RS Aggarwal Solutions Class 9 Chapter 8 Linear Equations in Two Variables. These Solutions are part of RS Aggarwal Solutions Class 9. we have given RS Aggarwal Solutions Class 9 Chapter 8 Linear Equations in Two Variables download pdf.

## RS Aggarwal Solutions Class 9 Chapter 8 Linear Equations in Two Variables

### Exercise 8A

Question 1:
(i) The given equation is x = 5
Take two solutions of the given equation as x = 5, y = 1 and x = 5, y = -1
Thus we get the following table: Plot points P(5,1) and Q(5,-1) on the graph paper.
Join PQ. The line PQ is the required graph. (ii) The given equation is y = -2
Take two solutions of the given equation as x = 1, y = -2 and x = 2, y = -2.
Thus we have the following table: Plot points P(1,-2) and Q(2,-2) on the graph paper. Join PQ. The line PQ is the required graph. (iii) The given equation is
x + 6 = 0
⇒ x = -6
Let x = -6 & y = 1
x = -6 & y = -1 Plot points P(-6,1) and Q(-6,-1) on the graph paper. Join PQ. The line PQ is the required graph. (iv) The given equation is
x + 7 = 0
⇒ x = -7
Let x = -7, y = 2 and x = -7, y = 1
Thus we have the following table: Plot points P(-7,2) and Q(-7,1) on the graph paper. Join PQ. The line PQ is the required graph. (v) y = 0 represents the x-axis
(vi) x = 0 represents the y-axis.

Question 2:
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: 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.

Question 3:
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: 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.

Question 4:
(i) The given equation is y = x
Let x = 1, then y = 1 and let x = 2, then y = 2
Thus, we have the following table: Plot points (1,1) and (2,2) on a graph paper and join them to get the required graph. (ii) The given equation is y = -x
Now, if x = 1, y = -1 and if x = 2, y = -2
Thus, we have the following table: Plot points (1,-1) and (2,-2) on a graph paper and join them to get the required graph. (iii) The given equation is y + 3x = 0
⇒ y = -3x
Now, if x = -1, then y = -3 (-1) = 3
And, if x = 1, then y = -3 (1) = -3
Thus we have the following table: Plot points (1,-3) and (-1,3) on a graph paper and join them to get the required graph. (iv) The given equation is 2x + 3y = 0
⇒ y = [latex s=2]frac { -2 }{ 3 }  [/latex] x
Now, if x = 3, then
y = [latex s=2]frac { -2 }{ 3 }  [/latex] × 3 = -2
And, if x = -3, then
y = [latex s=2]frac { -2 }{ 3 }  [/latex] × (-3) = 2
Thus, we have the following table Plot points (3,-2) and (-3,2) on a graph paper and join them to get the required graph. (v) The given equation is 3x – 2y = 0
⇒ y = [latex s=2]frac { 3 }{ 2 }  [/latex] x
Now, if x = 2,
y = [latex s=2]frac { 3 }{ 2 }  [/latex] × 2 = 3
And, if x = -2,
y = [latex s=2]frac { 3 }{ 2 }  [/latex] × (-2) = -3
Thus, we have the following table: Plot points (2,3) and (-2,-3) on a graph paper and join them to get the required graph. (vi) The given equation is 2x + y = 0
⇒ y = -2x
Now, if x = 1, then y = -2 1 = -2
And, if x = -1, then y = -2 (-1) = 2
Thus, we have the following table: Plot points (1,-2) and (-1,2) on a graph paper and join them to get the required graph. Question 5:
The given equation is, 2x – 3y = 5
⇒ y = [latex s=2]frac { 2x-5 }{ 3 }   [/latex]
Now, if x = 4, then
[latex s=2]y=frac { 2(4)-5 }{ 3 } =frac { 8-5 }{ 3 } =1   [/latex]
And, if x = -2, then
[latex s=2]y=frac { 2(-2)-5 }{ 3 } =frac { -4-5 }{ 3 } =frac { -9 }{ 3 } =-3   [/latex]
Thus, we have the following table: 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.

Question 6:
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: 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).

Question 7:
The given equation is 3x + 2y = 6
⇒ 2y = 6 – 3x
⇒ y = [latex s=2]frac { 6-3x }{ 2 }   [/latex]
Now, if x = 2, then
[latex s=2]y=frac { 6-3(2) }{ 2 } = 0   [/latex]
And, if x = 4, then
[latex s=2]y=frac { 6-3(4) }{ 2 } =frac { -6 }{ 2 } = -3   [/latex]
Thus, we have the following table: 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).

Complete RS Aggarwal Solutions Class 9

If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

Best of Luck For Your Future!!