#### Class Nine Mathematics NCERT Solution Chapter 4

**Chapter 4: Linear Equation in Two Variables**

**TEXTBOOK’S EXERCISE – 4.3 **

**1. **Draw a 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

**Sol. **(i) x + y = 4

For x = 0, y = 4

For x = 1, y = 3

So, we have the points (0, 4) and (1, 3).

(ii) x – y = 2

For x = 0, y = –2

For x = 1, y = –1

So, we have the points

(0, –2) and (1, –1).

(iii) y = 3x

For x = 0, y = 0

For x = 1, y = 3

So, we have the points

(0, 0) and (1, 3).

(iv) 3 = 2x + y

For x = 0, y = 3

x = 1, y = 1

So, we have the points

(0, 3) and (1, 1).

We get four lines, each one from (i), (ii), (iii) and (iv) when we join the points thus obtained.

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

**Sol. **Let x + y = k be such a
line, then

2 + 14 = k ⇒ k = 16.

∴ x + y = 16 passes through (2, 14).

Let 2x + 3y = k′ be another line through (2, 14).

2 × 2 + 3 × 14 = k′ ⇒ k′ = 4 + 42 = 46

⇒ 2x + 3y = 46 passes through (2, 14).

There are infinitely many such lines, as through a point an infinite number of straight 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.

**Sol. **(3, 4) lies on 3y = ax + 7

Therefore, substituting 3 for x and 4 for y in the above equation, we have 3 × 4 = a × 3 + 7 ⇒ 3a + 7 = 12 ⇒ 3a = 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.

**Sol. **Total fare y =
1 × 8 + (x – 1) 5,

where x is in km and y in Rs.

So, the required equation is y = 5 (x – 1) + 8

Now, for x = 1, y = 8

For, x = 2, y = 5 + 8 = 13

So, the points are (1, 8) and (2, 13).

Joining these points, we get the graph of the linear equation y = 5(x – 1) + 8.

**5. **From
the choices given below, choose the equation whose graphs are given in Fig. A
and Fig. B.

For Fig. A For Fig. B

(i) y = x (i) y = x + 2

(ii) x + y = 0 (ii) y = x – 2

(iii) y = 2x (iii) y = – x + 2

(iv) 2 + 3y = 7x (iv) x + 2y = 6

**Sol. **For fig. A

(ii) x + y = 0 is the correct choice.

This can be verified by putting

x = 1, y = –1 in x + y = 0

For fig. B

(iii) y = – x + 2 is the correct choice.

This can be verified by putting x = – 1, y = 3 in y = – x + 2