Solve the following system of equations: $\;$ $x^2 + y^2 + 6x + 2y = 0, \;\; x + y + 8 = 0$
Given system of equations:
$x^2 + y^2 + 6x + 2y = 0$ $\;\;\; \cdots \; (1)$
$x + y = 8$ $\;$ i.e. $\;$ $y = x - 8$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$x^2 + \left(-x - 8\right)^2 + 6x + 2 \left(-x - 8\right) = 0$
i.e. $\;$ $x^2 + x^2 + 16x + 64 + 6x - 2x - 16 = 0$
i.e. $\;$ $2x^2 + 20x + 48 = 0$
i.e. $\;$ $\left(x + 6\right) \left(x + 4\right) = 0$
i.e. $\;$ $x = -6$ $\;$ or $\;$ $x = -4$
Substituting the value of $x$ in equation $(2)$ gives
when $\;$ $x = -6$, $\;$ $y = - \left(-6\right) - 8 = -2$
and when $\;$ $x = -4$, $\;$ $y = - \left(-4\right) - 8 = -4$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-6, -2\right), \left(-4, -4\right) \right\}$