Linear Equations

Exercise 4.2

Question 1: Which one of the following statements is true and why? y = 3x + 5 has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions.

Solution: Since; y = 3x + 5 is a linear equation in two variables. We know that a linear equation in two variables has infinite solutions.
Hence option (iii) is true.

Question 2: Write four solutions for each of the following equations:

Solution: (i) 2x + y = 7
Or, y = 7 – 2x …………equation (1)
When x = 0, Putting the value in equation (1)
y = 7 – 2 xx 0
Or, y = 7 – 0
Or, y = 7

Putting the value x = 1 in equation (1)
y = 7 – 2 xx 1
Or, y = 7 – 2
Or, y = 5

Putting the value x = 2 in equation (1)
y = 7 – 2 xx 2
Or, y = 7 - 4
Or, y = 3

Putting the value x = 3 in equation (1)
y = 7 - 2 xx 3
Or, y = 7 - 6
Or, y = 1
Hence, four solutions for equation
2x + y = 7 are (0,7), (1,5), (2,3), (3,1).

(ii) π x + y = 9
Or, y = 9 - πx ……………… (1)
When x = 0, putting the value in equation (1)
y = 9 - π0
Or, y = 9 – 0 = 9
Putting the value x = 1 in equation (1)
Y = 9 - π
Putting the value x = - 1 in equation (1)
y = 9 + π
Putting the value x = 2 in equation (1)
y = 9 - 2π
Hence, four solutions for the given equation are; (0, 9), (1, 9 - π), (- 1, 9 + π), (2, 9 - 2π)

(iii) x = 4y …………equation (1)
When y = 0, Putting the value in equation (1)
x = 4 xx 0
Or, x = 0

Putting the value y = 1 in equation (1)
x = 4 xx 1
Or, x = 4

Putting the value y = -1 in equation (1)
x = 4 xx -1
Or, x = -4

Putting the value y = 2 in equation (1)
x = 4 xx 2
Or, x = 8
Hence, four solutions for equation x = 4y are (0, 0), (1, 4), (-4,-1), (8, 2).

Question 3: Check which of the following are solutions of the equation x – 2y = 4 and which are not:

Solution: (i) (0, 2)
The given equation is x – 2y = 4
L.H.S = x – 2y
Putting the value x = 0 and y = 2 to verify the solution (0, 2)
= 0 – 2 xx 2 = 0 – 4
= -4 ≠ R.H.S.
So, (0, 2) is not a solution of the equation x – 2y = 4

(ii) (2, 0)

Solution: The given equation is; x – 2y = 4
L.H.S = x – 2y
Putting the value x = 2 and y = 0 to verify the solution (2, 0)
= 2 – 2 xx 0 = 2 – 0
= 2 ≠ R.H.S.

So, (2, 0) is not a solution of the equation x – 2y = 4

(iii) (4, 0)

Solution: The given equation is; x – 2y = 4
L.H.S = x – 2y
Putting the value x = 4 and y = 0 to verify the solution (4, 0)
= 4 – 2 xx 0 = 4 – 0
= 4 = R.H.S.
So, (4, 0) is a solution of the equation x – 2y = 4.

(iv) (sqrt2, 4sqrt2)
Putting the value x = sqrt2 and y = 4sqrt2 to verify the solution:
= sqrt2 - 2 xx 4sqrt2
= sqrt2 - 8sqrt2
= 7sqrt2 ≠ RHS
Or, (sqrt2, 4sqrt2) is not a solution for the equation x – 2y = 4

(v) (1, 1)

Solution: The given equation is; x – 2y = 4
L.H.S = x – 2y
Putting the value x = 1 and y = 1 to verify the solution (1, 1)
= 1 – 2 xx 1 = 1 – 2 = -1 ≠ R.H.S.
So, (1, 1) is not a solution of the equation x – 2y = 4

Question 4: Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k

Solution: The given equation is; 2x + 3y = k
Putting the given value x = 2 and y = 1 in the equation
2 xx 2 + 3 xx 1 = k
4 + 3 = k
7 = k
Hence, k = 7