Question

A relation R on the set {0,1,2,…,10} defined by 

2x + 3y = 12.

Answer 1

Given, 

(x, y) R 2x + 3y = 12 

where x and y {0,1,2,…,10} 

2x + 3y = 12 

⇒ 2x = 12 – 3y

 ⇒ x = \(\frac{12-3y}{2}\) 

On putting y = 0,

⇒ x = \(\frac{12-3(0)}{2}\)

= \(\frac{12}{2}\) 

= 6

On putting y = 2,

⇒ x = \(\frac{12-3(2)}{2}\)

= \(\frac{12-6}{2}\)

= \(\frac{6}{2}\)

= 3

On putting y = 4,

⇒ x = \(\frac{12-3(4)}{2}\)

= \(\frac{12-12}{2}\)

= \(\frac{0}{2}\)

= 0

∴ R = {(0, 4), (3, 2), (6, 0)}