# Linear Equations

## Exercise 4.3 Part 2

Question 6: 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 is 5 units. Read from the graph the work done when the distance traveled by the body is

(i) 2 units (ii) 0 units

Solution: Let x units be the distance traveled by the body and y units be the work done by constant force.

According to problem, y = 5x …………….equation (1)
Now , putting the value x = 0 in equation (1)
y = 5 xx 0 = 0

So the solution is (0, 0)
Putting the value x = 1 in equation (1)
y = 5 xx 1 = 5
So the solution is (1, 5)

Putting the value x = 2 in equation (1)
y = 5 xx 2 = 10
So the solution is (2, 10)

So, the table of the different solutions of the equation is

(i) When x = 2 units (distance)
Putting the value x in equation (1)
y = 5x
y = 5 xx 2 = 10.
Hence, Work done = 10.

(ii) When x = 0 units (distance)
Putting the value x in equation (1)
y = 5x
y = 5 xx 0 = 0
Hence, Work done = 0.

Question 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 this data satisfies. (You may take their contributions as Rs. x and Rs. y). Draw the graph of the same.

Solution: Let the contribution of Yamini be x and that of Fatima be y.
According to problem, x + y = 100 …………..(1)
Now , putting the value x = 0 in equation (1)
0 + y = 100
y = 100. So the solution is (0, 100)
Putting the value x = 50 in equation (1)
50 + y = 100
y = 100 – 50
y = 50.
So the solution is (50, 50)
Putting the value x = 100 in equation (1)
100 + y = 100
Or, y = 100 – 100
y = 0
So the solution is (100, 0)
So, the table of the different solutions of the equation is

Question 8: In countries like the 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.

Solution: Given equation is

F=(9/5)C+32

(i) We have to take Celsius along x-axis and Fahrenheit along y-axis
Let C be x and F be y

So, the equation will be
y=(9/5)x+32 -----equation (1)

Now, putting the value x=0 in equation (1)
Or, y=(9/5)xx0+32=0+32=32

So, the solution is (0, 32)
Putting the value x=5 in equation (1)
Or, y=(9/5)xx5+32=9+32=41

So, the solution is (5, 41)
Putting the value x=-5 in equation (1)
Or, y=(9/5)xx(-5)+32=-9+32=23

So, the solution is (-5, 23)

So, the table of the different solutions of the equation is

(ii) When C = 30°,
F=(9/5)xx30+32=54+32=86°

(iii) When F = 95°,
95=(9/5)xxC+32

Or,95-32=(9/5)C

Or, 63=(9/5)C

Or, C=63xx(5/9)=35

(iv) When C = 0,
F=(9/5)xx0+32=0+32=32

When x° F = x° C
F=(9/5)C+32

Or, x=(9/5)x+32

Or, x-(9/5)x=32

Or, (5x-9x)/5=32

Or, -4x=32xx5=160
Or, x=160/-4=-40°

## Exercise 4.4

Question 1: Give the geometric representations of y = 3 as an equation.

(i) in one variable (ii) in two variables.

Solution: (i) y = 3
(ii) y = 3
Or, 0x + y = 3
Or, 0x + y – 3 = 0
which is in fact y = 3
It is a line parallel to x-axis at a positive distance of 3 from it. We have two solution for it. i.e. (0, 3), (1, 3).

Question 2: Give the geometric representation of 2x + 9 = 0 as and equation,
(i) in one variable (ii) in two variables.

Answer: (i) 2x + 9 = 0
Or, 2x = -9Or, x = - 9/2 = - 4.5

(ii) Given equation is
2x + 9 = 0
2x + 0y + 9 = 0 {we know that it is actually 2x + 9 = 0}
Or, x = - 9/2 = - 4.5
It is line parallel to y-axis at a negative distance we have the two points lying it, the points are A(-4.5, 0), B(-4.5, 2).