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

3x + 4y = 12, (a + b) x + 2 (a – b) y = 5a – 1

Given, pair of equations

3x + 4y = 12

and (a + b)x + 2(a – b)y = 5a – 1

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

a₁ = 3, b₁ = 4 and c₁ = – 12

and a_{2 =} (a + b), b₂ = 2(a – b) and c₂ = – (5a – 1) = 1 + 5a

and

For infinitely many solutions,

Here,

∴

On taking I and II terms, we get

⇒ 6(a – b) = 4(a + b)

⇒ 6a – 6b = 4a + 4b

⇒ 6a – 4a – 6b – 4b = 0

⇒ 2a – 10b = 0

⇒ a – 5b = 0 …(1)

On taking I and III terms, we get

⇒ 3(5a – 1) = 12(a + b)

⇒ 15a – 3 = 12a + 12b

⇒ 15a – 12a – 12b = 3

⇒ 3a – 12b = 3

⇒ a – 4b = 1 …(2)

Solving eqⁿ (1) and (2), we get

⇒ b = 1

Now, substituting the value of b in eqⁿ (2), we get

⇒ a – 4b = 1

⇒ a – 4 = 1

⇒ a = 1 + 4

⇒ a = 5