Find four solutions for the following equation: 2(x+3)-3(y+1)=0

2(x+3)-3(y+1) = 0
          Take x = 1,
2(1+3)-3(y+1) = 0    ⇒ 2× 4-3y-3 = 0 ⇒ -3y = 0-8+3 = -5                ⇒ y =                ∴ y =
          Take x = 2,
2(2+3)-3(y+1) = 0     Þ 2× 5-3y-3 = 0       ⇒ 10-3y-3 = 0              ⇒ -3y = -10+3
                    = -7                 ∴y ==
Take x = 3,
2(3+3)-3(y+1) = 0    ⇒ 2× 6-3y-3 = 0
Þ -3y=-12+3 =-9               ⇒ y =
                     = 3
          Take x = 4,
2(4+3)-3(y+1) = 0    ⇒ 2× 7-3y-3 = 0             ⇒ -3y = -14+3
                       = -11        ∴ y = = ∴ The four solutions of the given equation are x = 1, y = ; x = 2, y = , x = 3, y = 3 and x = 4, y = .