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## 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).

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