Solve the following system of equations: $\;$ $x^2 - xy = 28, \;\; y^2 - xy = -12$
Given system of equations:
$x^2 - xy = 28$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 - xy = -12$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ gives
$x^2 + y^2 - 2xy = 16$
i.e. $\;$ $\left(x - y\right)^2 = 16$ $\implies$ $x - y = \pm 4$
i.e. $\;$ $x = y + 4$ $\;\;\; \cdots \; (3)$ $\;\;$ or $\;\;$ $x = y - 4$ $\;\;\; \cdots \; (4)$
In view of equation $(3)$, equation $(1)$ becomes
$\left(y + 4\right)^2 - \left(y + 4\right)y = 28$
i.e. $\;$ $y^2 + 8y + 16 - y^2 - 4y = 28$
i.e. $\;$ $4y = 12$ $\implies$ $y = 3$
Substituting $y = 3$ in equation $(3)$ gives $\;\;$ $x = 3 + 4 = 7$
In view of equation $(4)$, equation $(1)$ becomes
$\left(y - 4\right)^2 - \left(y - 4\right)y = 28$
i.e. $\;$ $y^2 - 8y + 16 - y^2 + 4y = 28$
i.e. $\;$ $-4y = 12$ $\implies$ $y = -3$
Substituting $y = -3$ in equation $(4)$ gives $\;\;$ $x = -3 - 4 = -7$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(7, 3\right), \; \left(-7, -3\right) \right\}$