Given that,
(a – b) x + (a + b) y = 2a2 – 2b2
(a + b) (x + y) = 4ab
(a – b) x + (a + b) y – 2(a2 – b2) = 0
(a + b)x + (a + b)y – 4ab = 0
On comparing both the equation with the general form we get
a1 = a – b, b1 = a + b, c1 = -2,
a2 = a + b, b2 = a + b, c2 = -4ab
Now by using cross multiplication we get
x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)
⇒ x/(-(a + b)4ab + 2(a + b) (a2 – b2)) = y/(− 2(a2 − b2)(a + b) + 4ab(a – b))
= 1/((a − b)(a + b) − (a + b)(a + b))
⇒ x/(2(a + b)(a2 – b2 + 2ab)) = 1/-2b(a + b)
x = (2ab – a2 + b2)/b
and,
= -y/(2(a – b) (a2 + b2) -2b (a + b)) = 1/ -2b(a + b)
y = (a – b)(a2 + b2)/ b(a + b)
Hence, x = (2ab – a2 + b2)/b and y = (a – b)(a2 + b2)/ b(a + b)