Solve the following system of equations: $\;$ $x^2 + xy = 15, \;\; y^2 + xy = 10$
Given system of equations:
$x^2 + xy = 15$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 + xy = 10$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ gives
$x^2 + y^2 + 2xy = 25$
i.e. $\;$ $\left(x + y\right)^2 = 25$ $\implies$ $x + y = \pm 5$
i.e. $\;$ $x = 5 - y$ $\;\;\; \cdots \; (3)$ $\;\;$ or $\;\;$ $x = -5 - y$ $\;\;\; \cdots \; (4)$
When $\;$ $x = 5 - y$, $\;$ equation $(1)$ becomes
$\left(5 - y\right)^2 + \left(5 - y\right)y = 15$
i.e. $\;$ $25 + y^2 - 10y + 5y - y^2 = 15$
i.e. $\;$ $5y = 10$ $\implies$ $y = 2$
Substituting $y = 2$ in equation $(3)$ gives $\;\;$ $x = 5 - 2 = 3$
When $\;$ $x = -5 - y$, $\;$ equation $(1)$ becomes
$\left(-5 - y\right)^2 - \left(5 + y\right)y = 15$
i.e. $\;$ $25 + y^2 + 10y - 5y - y^2 = 15$
i.e. $\;$ $5y = -10$ $\implies$ $y = -2$
Substituting $y = -2$ in equation $(4)$ gives $\;\;$ $x = -5 + 2 = -3$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(3, 2\right), \; \left(-3, -2\right) \right\}$