Question

$$\begin{gathered} (a-b)x+(a+b)y=a^2-2ab-b^2, \\ (a+b)(x+y)=a^2+b^2. \end{gathered}$$

Answer 1

The given equations are:
(a − b)x + (a + b)y = a² − 2ab − b² ...(i)
(a + b)(x + y) = a² + b² ...(ii)
From (ii), we have:
(a + b)x + (a + b)y = a² + b² ...(iii)
On subtracting (iii) from (i), we get:
(a − b − a − b)x = (a² − 2ab − b² − a² − b² )
⇒ −2bx = −2ab − 2b²
⇒ −2bx = −2b(a + b)
⇒ x = (a + b)
On substituting x = (a + b) in (i), we get:
(a − b)(a + b) + (a + b)y = a² − 2ab − b²
⇒ a² − b² + (a + b)y = a² − 2ab − b²
⇒ (a + b)y = −2ab
⇒ $y=\frac{-2ab}{\left(a+b\right)}$
Hence, the required solution is $x=\left(a+b\right)$ and $y=\frac{-2ab}{\left(a+b\right)}$.