Question

Solve the following equations : 

i. 4(x + 12) = 8 

ii. 3y + 4 = 5y – 6

Answer 1

i. 4(x + 12) = 8 

∴ 4x + 48 = 8 

∴ 4x + 48 – 48 = 8 – 48

….(Subtracting 48 from both sides) 

∴ 4x = -40

∴ \(x = \frac{-40}{4}\)

∴ x = -10

ii. 3y + 4 = 5y – 6 

∴ 3y + 4 + 6 = 5y – 6 + 6 

….(Adding 6 on both sides) 

∴ 3y + 10 = 5y 

∴ 3y + 10 – 3y = 5y – 3y 

….(Subtracting 3y from both sides) 

∴ 10 = 2y 

∴ 2y = 10

∴ y = 10/2

∴ y = 5