(a+ b)x+(a-b)y=a^2+b^2 (a-b)x+(a+b)y=a^2 + b^2

The given system of linear equation is
(a – b)x + (a + b)y = a² – 2ab –  b² ....(1)
(a + b)(x + y) = a² + b² ....(2)
Equation (2) can be written as
(a + b)x + (a + b)y = a² + b² ....(3)
Subtracting equation (3) from equation (1), we get
– 2bx = – 2ab – 2b²
{tex}\Rightarrow{/tex} – 2bx = – 2b(a + b)
{tex}\Rightarrow{/tex} x = a + b
Substituting this value of x in equation (1), we get
(a – b)(a + b)(a + b)y = a²  – 2ab – b²
{tex}\Rightarrow{/tex} a² – b² + (a + b)y = a² – 2ab – b²
{tex}\Rightarrow{/tex} (a + b)y = – 2ab
{tex}\Rightarrow \;y = - \frac{{2ab}}{{a + b}}{/tex}
Verification: Substituting x = a + b, {tex}y = - \frac{{2ab}}{{a + b}}{/tex},
We find that both the equations (1) and (2) are satisfied as shown below:
(a – b)x + (a + b)y = (a – b)(a + b) + (a + b) {tex}\left\{ { - \frac{{2ab}}{{a + b}}} \right\}{/tex}
= a² – 2ab – b²
(a + b)(x + y) = (a + b){tex}\left\{ {(a + b) + (- \frac{{2ab}}{{a + b}})} \right\}{/tex}
= (a + b)² – 2ab
= a² + b²
hence, the solution we have got is correct.