# Chapter 4 Linear Equations in Two Variables RS Aggarwal Solution for Class 9th Maths

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LinearEquations in Two Variables  Exercise 8A

# Question 1:

1. 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. 1. 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.

1. 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. 1. 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. 1. y = 0 represents the x-axis
2. 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:

1. 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. 1. 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.

1. 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.

1. The given equation is 2x + 3y = 0

⇒ y = x

Now, if x = 3, then y = × 3 = -2

And, if x = -3, then y = × (-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. 1. The given equation is 3x – 2y = 0

⇒ y = x Now, if x = 2, y = × 2 = 3

And, if x = -2,

y = × (-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. 1. 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 =  Now, if x = 4, then

And, if x = -2, then  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. 1. 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.

1. 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 =  Now, if x = 2, then

And, if x = 4, then  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).