Find two solutions for each of the following:

(i) 3x + 4y = 12 (ii) 3x + 5y = 0

(iii) 4y + 5 = 0

(i) 3x + 4y = 12

⇒ 4y = 12 – 3x

Let x = 4,

⇒ y = 0

Therefore, (4, 0) is a solution

Let x = – 4,

⇒ y = 6

Therefore, ( – 4, 6) is a solution

(ii) 3x + 5y = 0

⇒ 5y = 0 – 3x

Let x = 5,

⇒ y = – 3

Therefore, (5, – 3) is a solution

Let x = – 5,

⇒ y = 3

Therefore, ( – 5, 3) is a solution

(iii) 4y + 5 = 0

⇒ 4y = 0 – 5

Let x = 1,

Therefore, is a solution.

Let x = – 1,

Therefore, is a solution.