(3a + 1) x + 3y – 2 = 0, (a^2 +1) x + (a – 2) y – 5 = 0

Question

For what value of a, the following system of linear equations has no solutions:

(3a + 1) x + 3y – 2 = 0, (a² +1) x + (a – 2) y – 5 = 0

Answer 1

iven, pair of equations

(3a + 1)x + 3y – 2 = 0

and (a² + 1)x + (a – 2)y – 5 = 0

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

a₁ = 3a + 1, b₁ = 3 and c₁ = – 2

and a_{2 =} a² + 1, b₂ = a – 2 and c₂ = – 5

For no solutions,

⇒ (3a + 1)(a – 2) = 3(a² + 1)

⇒ 3a² – 6a + a – 2 = 3a² + 3

⇒ – 5a = 2 + 3

⇒ a = – 1