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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

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