# NCERT Solutions For Class 9 Mathematics | Linear Equations In Two Variables | Chapter 4 Exercise 4.3

1. Draw the graph of each of the following linear equations in two variables:

i) x + y = 4

ii) x – y = 2

iii) y = 3x

iv) 3 = 2x + y

Answer:

Given equation is x + y = 4

We need to find the two points.

When x = 0, y = 4

When y= 4, x = 4 – y = 4 – 4 = 0

Therefore, two points are (0, 4) and (4, 0).

Given equation is x – y =2

When x = 2, 2 – y = 2 then y= 0

When x = 0, 0 – y = 2 then y = -2

Therefore, two points are (2, 0) and (0,-2).

Given equation is y = 3x

When x =1, y = 3

When x = -1, y = -3

Therefore, the two points are (1, 3) and (-1,-3).

Given equation is 3 = 2x + y

When x = 0, 3 = 2×0 + y then y =3

When y = 0, 3 = 2x + 0 then x =3/2.

Therefore, the two points are (0, 3) and (3/2, 0).

2. Give the equations of two lines passing through (2, 14). How many more such lines are there and why?

Answer:

When x= 2 and y= 14, one equation is x + y = 2 + 14 = 16

Another equation is y = 7x or 7x – y = 0

Therefore, 7x – y = 0 and x + y = 16 are the two equations of two lines passing through (2, 14).

Infinitely many such lines are there, because through a point infinitely many lines can be drawn.

3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?

Answer:

Given 3y = ax + 7

If the point (3, 4) lies on the graph of the equation 3y = ax +7,

we can take x =3 and y = 4.

Then, 3 (4) = a (3) +7

12 = 3a + 7

3a = 12 – 7= 5

a = 5/3.

4. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer:

Take the distance covered = x km and total fare = Rs y.

Fare for the first kilometer = Rs 8

Fare for the subsequent distance = Rs 5

Fare for the rest of the distance = Rs (x-1) 5

The linear equation for this information is given by,

y = 8 + (x -1)5

y = 8 + 5x – 5

y = 3 + 5x

5x – y +3 =0

When x = 0, y = 3

When x = 1, y = 3 + 5 x 1 = 8

Therefore, the two points are (0, 3) and (1, 8).