# NCERT Solutions For Class 9 Math Chapter – 4 Exercise – 4.3

NCERT Solutions For Class 9 Math Chapter – 4 Exercise – 4.3

Q1. Draw the graph of each of the following linear equations in tow variables:

• X+y = 4
• X – y = 2
• Y = 3x
• 3 = 2x + y

Solution:-

• x + y = 4

Put x = 0  then y = 4

Put x = 4 then y = 0

 X 0 4 Y 4 0

• X – y = 2

Put x = 0 then y = -2

Put x = 2 then y = 0

 x 0 2 y -2 0

• Y = 3x

Put x = 0 then y = 0

Put x = 1 then y = 3

 x 0 1 y 0 3

• 3 = 2x + y

Put x = 0 then y = 3

Put x = 1 then y = 1

 x 0 1 y 3 1

Q2. Give the equations of two lines passing through How many more such lines are there , and why ?

Solution:-

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.

Q3. If the point (3,4) lies on the graph of the equation 3y = ax + 7 find the value of a

Solution:-

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 x 4 = a x 3 + 7

12 = 3a + 7

3a = 12 – 7

A = 5/3

Q4. The taxi fare in a city is as follows: For the first kilometer , 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.

Solution:-

Total fare = y

Total di-3/5stance covered = x

Y = 8 + 5(x-1)

Y = 8 + 5x – 5

Y = 5x + 3

 x 0 -3/5 y 3 0

Q5. From the choices given below , Choose the equation whose graphs are given in

• Y = x
• Y = x + 2
• X + y = 0
• Y = x – 2
• Y = 2x
• Y = -x + 2
• 2 + 3y = 7x
• X + 2y = 6

Solution:-

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.

Points are (-1,3) , (0,2) and (2,0).

Y = -x + 2 is correct as it satisfies all the value of the points.

Q6. If the work done by a body on application of a constant force is directly proportionally 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

• 2 units      (ii)  0 units

Solution:-

y = 5x

• When x = 2 units then y = 10 units
• When x = 0 unit then y = 0 unit
 x 2 0 y 10 0

Q7. Yamini and Fatima , two students of Class IX of a school , together contributed Rs 100 towards the Prime Minister Relief fund to help the earthquake victims Write a linear equation which satisfies this data Draw the graph of the same.

Solution:-

x + y = 100

When x= 0 then y= 100

When  x = 100 then y = 0

 x 0 50 100 y 100 50 0

Q8. In a countries like USA and Canada temperature is  measured in Fahrenheit , whereas in countries like India It is measured in Celsius Here a linear equation that converts Fahrenheit to Celsius:

F= (9/5)C + 32

• Draw the graph of the linear equation above using Celsius for x – axis and Fahrenheit for y – axis.
• If the temperature is 300C What is the temperature in Fahrenheit ?
• If the temperature is 950F What is the temperature in Celsius?
• If the temperature is 00C, What is the temperature in Fahrenheit and if the temperature is 00 F What is the temperature in Celsius?
• Is there a temperature which is numerically the same in both fahrenheit and Celsius? If yes , find it.

Solution:-

• F = (9/5)C + 32

When C = 0 then F =  32

Also , when C =  -10 then F = 14

 x 0 -10 y 32 14

• Putting the value of C = 30 in F = (9/5)C + 32 we get

F = 54 + 32

F = 86

• F = 95  in F = (9/5)C + 32 we get

95 = (9/5)C + 32

= (9/5)C = 95 – 32

C = 35

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

0 = (9/5)C + 32

C = -160/9

Putting value of C = 0 in F = (9/5)C + 32 WE GET

F = (9/5) X 0 + 32

F = 32

• Putting F = C in F = (9/5)C + 32 we get

F = (9/5)F + 32

F = -40