# Linear Equations

## Exercise 4.3

Question 1: Draw the graph of each of the following linear equation in two variables:

(i) x + y = 4 (ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y

Solution: (i) The given equation is
x + y = 4
y = 4 – x …………….equation (1)
Now , putting the value x = 0 in equation (1)
y = 4 – 0 = 4. So the solution is (0, 4)

Putting the value x = 1 in equation (1)
y = 4 – 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1)
y = 4 – 2 = 2. So the solution is (2, 2)
So, the table of the different solutions of the equation is

Solution:(ii) The given equation is
x - y = 2
x = 2 + y…………….equation (1)

Now , putting the value y = 0 in equation (1)
x = 2 + 0 = 2. So the solution is (2, 0)

Putting the value y = 1 in equation (1)
x = 2 + 1 = 3. So the solution is (3, 1)
Putting the value y = 2 in equation (1)
x = 2 + 2 = 4. So the solution is (4, 2)
So, the table of the different solutions of the equation is

Solution:(iii) The given equation is
y = 3x
y = 3x …………….equation (1)

Now , putting the value x = 0 in equation (1)
y = 3 x 0 = 0. So the solution is (0, 0)
Putting the value x = 1 in equation (1)
y = 3 x 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1)
y = 3 x 2 = 6. So the solution is (2, 6)
So, the table of the different solutions of the equation is

Solution:(iv) The given equation is
3 = 2 x + y
2 x + y = 3
y = 3 – 2x …………….equation (1)
Now, putting the value x = 0 in equation (1)
y = 3 – 2 x 0
y = 3 – 0 = 3. So the solution is (0, 3)
Putting the value x = 1 in equation (1)
y = 3 – 2 x 1
y = 3 – 2 = 1. So the solution is (1, 1)
Putting the value x = 2 in equation (1)
y = 3 – 2 x 2
y = 3 – 4 = -1. So the solution is (2, -1)
So, the table of the different solutions of the equation is

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

Solution: Since the given solution is (2, 14)
Therefore, x = 2 and y = 14
Or, One equation is x + y = 2 + 14 = 16
x + y = 16
Second equation is x – y = 2 – 14 = -12
x - y = -12
Third equation is y = 7x
0 = 7x – y
7x - y = 0

In fact we can find infinite equations because infinite number of lines pass through one point.

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

Solution: The given equation is
3y = ax + 7 ……………….equation (1)
According to problem, point (3, 4) lie on it.
So, putting the value x = 3 and y = 4 in equation (1)
3 xx 4 = a xx 3 + 7
12 = 3a + 7
12 – 7 = 3a
5 = 3a

Or, 5/3=a

Now, putting the value of ‘a’ in equation (1)
3y=(5/3)x+7

Now, for x=3 and y=4
LHS =3y=3xx4=12
RHS, (5/3)x+7=(5/3)xx3+7=5+7=12

Hence, LHS=RHS
Or, 3y=(5/3)x+7

Question 4: 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 kilometer. Taking the distance covered as x km and total fares as Rs. y, write a linear equation for this information, and draw its graph.

Solution: Given, Taxi fare for first kilometer = Rs. 8
Taxi fare for subsequent distance = Rs. 5
Total distance covered = x
Total fare = y
Since the fare for first kilometer = Rs. 8
According to problem, Fare for (x – 1) kilometer = 5(x-1)
So, the total fare = 5(x-1) + 8
y = 5(x-1) + 8
y = 5x – 5 + 8
y = 5x + 3
Hence, y = 5x + 3 is the required linear equation.
Now the equation is; y = 5x + 3 …………….equation (1)
Now, putting the value x = 0 in equation (1)
y = 5 x 0 + 3
y = 0 + 3 = 3 So the solution is (0, 3)
Putting the value x = 1 in equation (1)
y = 5 x 1 + 3
y = 5 + 3 = 8. So the solution is (1, 8)
Putting the value x = 2 in equation (1)
y = 5 x 2 + 3
y = 10 + 3 = 13. So the solution is (2, 13)
So, the table of the different solutions of the equation is

Question 5: From the choices given below, choose the equation whose graphs are given in Fig. 1 and Fig. 2.
(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

Solution: From the given Fig. 1, the solutions of the equation are (-1, 1), (0, 0) and (1, -1)
Therefore the equation which satisfies these solutions is the correct equation.
Equation (ii) x + y = 0, satisfies these solutions.

Proof: Putting the value x = -1 and y = 1 in the equation x + y = 0
L.H.S = x + y = -1 + 1 = 0 = R.H.S
Putting the value x = 0 and y = 0
L.H.S = x + y = 0 + 0 = 0 = R.H.S
Putting the value x = 1 and y = -1
L.H.S = x + y = 1 + (-1) = 1 – 1 = 0 = R.H.S

Hence, option (ii) x + y = 0 is correct.
From the given Fig. 2, the solutions of the equation are (-1, 3), (0, 2) and (2, 0)
Therefore the equation which satisfies these solutions is the correct equation.
Equation (i) y = -x + 2 , satisfies these solutions.

Proof: Putting the value x = -1 and y = 3 in the equation y = -x + 2
x + y = 2
L.H.S = x + y = -1 + 3 = 2 = R.H.S
Putting the value x = 0 and y = 2
L.H.S = x + y = 0 + 2 = 2 = R.H.S
Putting the value x = 2 and y = 0
L.H.S = x + y = 2 + 0 = 2 = R.H.S
Hence, option (i) y = -x + 2 is correct.