Question -
Answer -
(i) 3x + 4y =7 ….(1)
Step 1: Isolate above equation in y.
Subtract 3x from both the sides,
3x + 4y – 3x = 7 – 3x
4y = 7 – 3x
Divide each side by 4
y = 1/4 x (7 – 3x) ….(2)
Step 2: Find Solutions
Substituting x = 1 in (2)
y = 1/4 x (7 – 3) = 1/4 x 4 = 1
Thus x = 1 and y = 1 is the solution of 3x + 4y = 7
Again, Substituting x = 2 in (2)
y = 1/4 x (7 – 3 x 2) = 1/4 x 1 = 1/4
Thus x = 2 and y = 1/4 is the solution of 3x + 4y = 7
Therefore, (1, 1) and (2, 1/4) are two solution of 3x + 4y = 7.
(ii) Given: x = 6y
Substituting x =0 in the given equation,
0 = 6y
or y = 0
Thus (0,0) is one solution
Again, substituting x=6
6 = 6y
or y = 1
Thus, (6, 1) is another solution.
Therefore, (0, 0) and (6, 1) are two solutions of x = 6y.
(iii) Given: x + πy = 4
Substituting x = 0 ⇒ 0 + πy = 4 ⇒ y = 4/ π
Substituting y = 0 ⇒ x + 0 = 4 ⇒ x = 4
Therefore, (0, 4/ π) and (4, 0) are two solutions of x + πy = 4.
(iv) Given: 2/3 x – y = 4
Substituting x = 0 ⇒ 0 – y = 4 ⇒ y = -4
Substituting x = 3 ⇒ 2/3 × 3 – y = 4 ⇒ 2 – y = 4 ⇒ y = -2
Therefore, (0, -4) and (3, -2) are two solutions of 2/3 x – y = 4.