Solve the following system of equations: $\;$ $x^2 - y^2 = 16, \;\; x + y = 8$
Given system of equations:
$x^2 - y^2 = 16$ $\;$ i.e. $\;$ $\left(x + y\right) \left(x - y\right) = 16$ $\;\;\; \cdots \; (1)$
$x + y = 8$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$8 \left(x - y\right) = 16$
i.e. $\;$ $x - y = 2$ $\;\;\; \cdots \; (3)$
Adding equations $(2)$ and $(3)$ gives
$2x = 10$ $\implies$ $x = 5$
Substituting $x = 5$ in equation $(2)$ gives
$5 + y = 8$ $\implies$ $y = 3$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(5, 3\right) \right\}$