# 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

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,

a1 = (a + b),

b1 = (a — b),

c1 = – 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).

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