Solve the following pair of linear equations by the method of cross – multiplication:

(a + b)x + (a — b)y = a² + 2ab — b², a ≠ b

(a — b) (x + y) = a² — b², a ≠ b (CBSE 2004, 2007, 2008)

Given equations are

(a + b)x + (a — b)y = a² + 2ab — b²

And (a — b) (x + y) = a² — b²

Change the given equations to the form,

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 …(i)

⇒ (a + b)x + (a — b)y – a² – 2ab + b² = 0

And

(a – b)x + (a – b)y – a² + b² = 0

On comparing with (i) we get,

a_{1 =} (a + b),

b₁ = (a — b),

c_{1 =} – a² – 2ab + b²;

a₂ = (a – b),

b₂ = (a – b),

c₂ = – a² + b² = b² – a² = (b + a)(b – a) {Using a² – b² = (a + b)(a – b)}

Applying cross multiplication method which says,

Putting the given values in the above equation we get,

Similarly,

∴The solution of the pair of equations is (a, b).