Solve the following system of equations: $\;$ $2y^2 + xy -x^2 = 0, \;\; x^2 - xy - y^2 + 3x + 7y + 3 = 0$
Given system of equations:
$2y^2 + xy -x^2 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^2 - xy - y^2 + 3x + 7y + 3 = 0$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$2y^2 + 2xy -xy - x^2 = 0$
i.e. $\;$ $2y \left(y + x\right) - x \left(y + x\right) = 0$
i.e. $\;$ $\left(2y - x\right) \left(y + x\right) = 0$
i.e. $\;$ $x = 2y$ $\;\;$ or $\;\;$ $x = -y$
Case 1:
When $\;$ $x = 2y$ $\;\;\; \cdots \; (3)$, $\;$ equation $(2)$ becomes
$\left(2y\right)^2 - 2y \times y - y^2 + 3 \times 2y + 7y + 3 = 0$
i.e. $\;$ $4y^2 - 2y^2 - y^2 + 6y + 7y + 3 = 0$
i.e. $\;$ $y^2 + 13y + 3 = 0$
i.e. $\;$ $y = \dfrac{-13 \pm \sqrt{13^2 - 4 \times 1 \times 3}}{2} = \dfrac{-13 \pm \sqrt{157}}{2}$
When $\;$ $y = \dfrac{-13 + \sqrt{157}}{2}$, $\;$ we have from equation $(3)$,
$x = 2 \times \left(\dfrac{-13 + \sqrt{157}}{2}\right) = -13 + \sqrt{157}$
and when $\;$ $y = \dfrac{-13 - \sqrt{157}}{2}$, $\;$ we have from equation $(3)$,
$x = 2 \times \left(\dfrac{-13 - \sqrt{157}}{2}\right) = -13 - \sqrt{157}$
Case 2:
When $\;$ $x = -y$ $\;\;\; \cdots \; (4)$, $\;$ equation $(2)$ becomes
$\left(-y\right)^2 - \left(-y\right) y - y^2 + 3 \left(-y\right) + 7y + 3 = 0$
i.e. $\;$ $y^2 + y^2 - y^2 - 3y + 7y + 3 = 0$
i.e. $\;$ $y^2 + 4y + 3 = 0$
i.e. $\;$ $\left(y + 3\right) \left(y + 1\right) = 0$
i.e. $\;$ $y = -3$ $\;\;$ or $\;\;$ $y = -1$
When $\;$ $y = -3$, $\;$ we have from equation $(4)$, $\;\;$ $x = - \left(-3\right) = 3$
and when $\;$ $y = -1$, $\;$ we have from equation $(4)$, $\;\;$ $x = - \left(-1\right) = 1$
$\therefore \;$ The solution to the given system of equations is
$\left(x, y\right) = \left(-13 + \sqrt{157}, \dfrac{-13 + \sqrt{157}}{2}\right), \; \left(-13 - \sqrt{157}, \dfrac{-13 - \sqrt{157}}{2}\right)$,
$\hspace{2cm} \left(3, -3\right), \; \left(1, -1\right)$