Solve the following system of equations: $\;$ $2x^2 + y^2 + 3xy = 12, \;\; 2 \left(x + y\right)^2 - y^2 = 14$
Given system of equations:
$2x^2 + y^2 + 3xy = 12$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $2 \left(x + y\right)^2 - y^2 = 14$ $\;\;\; \cdots \; (2)$
Equation $(2)$ can be written as
$2x^2 + 4xy + 2y^2 - y^2 = 14$
i.e. $\;$ $2x^2 + 4xy + y^2 = 14$ $\;\;\; \cdots \; (3)$
Subtracting equations $(1)$ and $(3)$ gives
$x \; y = 2$ $\;\;$ i.e. $\;$ $x = \dfrac{2}{y}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(1)$ becomes
$2 \times \left(\dfrac{2}{y}\right)^2 + y^2 + 3 \times \dfrac{2}{y} \times y = 12$
i.e. $\;$ $\dfrac{8}{y^2} + y^2 + 6 = 12$
i.e. $\;$ $y^4 - 6y^2 + 8 = 0$
i.e. $\;$ $\left(y^2 - 2\right) \left(y^2 - 4\right) = 0$
i.e. $\;$ $y^2 = 2$ $\;\;$ or $\;\;$ $y^2 = 4$
i.e. $\;$ $y = \pm \sqrt{2}$ $\;\;$ or $\;\;$ $y = \pm 2$
When $\;$ $y = \pm \sqrt{2}$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{2}{\pm \sqrt{2}} = \pm \sqrt{2}$
When $\;$ $y = \pm 2$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{2}{\pm 2} = \pm 1$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\sqrt{2}, \sqrt{2}\right), \; \left(- \sqrt{2}, - \sqrt{2}\right), \; \left(1, 2\right), \; \left(-1, -2\right) \right\}$