(a – 1) x + 3y = 2, 6x + (1 – 2b) y = 6

Question

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

(a – 1) x + 3y = 2, 6x + (1 – 2b) y = 6

Answer 1

Given, pair of equations

(a – 1)x + 3y = 2

and 6x + (1 – 2b)y = 6

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = (a – 1), b₁ = 3 and c₁ = – 2

and a_{2 =} 6, b₂ = 1 – 2b and c₂ = – 6

For infinitely many solutions,

⇒ 3(a – 1) = 6

⇒ 3a – 3 = 6

⇒ 3a = 6 + 3

⇒ 9 = 1 – 2b

⇒ – 2b = 9 – 1

⇒ – 2b = 8

⇒ b = – 4