# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 4 LINEAR EQUATIONS IN TWO VARIABLES EX – 4.3

## Linear Equations in Two Variables

Page No: 74

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

(i) x + y = 4
Put x = 0 then y = 4
Put x = 4 then y = 0

(ii) x – y = 2
Put x = 0 then y = -2
Put x = 2 then y = 0

(iii) y = 3x
Put x = 0 then y = 0
Put x = 1 then y = 3

(iv) 3 = 2x + y
Put x = 0 then y = 3
Put x = 1 then y = 1

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

Here, x = 2 and y =14.
Thus, x + y = 1
also, y = 7x ⇒ y – 7x = 0
∴ The equations of two lines passing through (2, 14) are
x + y = 1 and y – 7x = 0.
There will be infinite such lines because infinite number of lines can pass through a given point.

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

The point (3, 4) lies on the graph of the equation.
∴ Putting x = 3 and y = 4 in the equation 3y = ax + 7, we get
3×4 = a×3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7
⇒ 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.

Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3

5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.

In fig. 4.6, Points are (0, 0), (-1, 1) and (1, -1).
∴ Equation (ii) x + y = 0 is correct as it satisfies all the value of the points.

In fig. 4.7, Points are (-1, 3), (0, 2) and (2, 0).
∴ Equation (iii) y = –x + 2 is correct as it satisfies all the value of the points.

Page No: 75

6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is

(i) 2 units               (ii) 0 unit

Let the distance traveled by the body be x and y be the work done by the force.
y ∝ x (Given)
⇒ y = 5x (To equate the proportional, we need a constant. Here, it was given 5)
A/q,
(i) When x = 2 units then y = 10 units
(ii) When x = 0 unit then y = 0 unit

7. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Let the contribution amount by Yamini be x and contribution amount by Fatima be y.
A/q,
x + y = 100
When x = 0 then y = 100
When x = 50 then y = 50
When x = 100 then y = 0

8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:  F = (9/5)C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

(i) F = (9/5)C + 32When C = 0 then F = 32also, when C = -10 then F = 14

(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30  + 32
⇒ F = 54 + 32
⇒ F = 86

(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C  + 32
⇒ (9/5)C = 95 – 32
⇒ C = 63 × 5/9
⇒ C = 35

(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C  + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9

Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0  + 32
⇒ F = 32

(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F – 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.