Solve the following system of equations: $\;$ $4x^2 + y^2 - 2xy = 7, \;\; \left(2x - y\right) y = y$
Given system of equations:
$4x^2 + y^2 - 2xy = 7$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $\left(2x - y\right) y = y$ $\;\;\; \cdots \; (2)$
Case 1:
When $\;$ $y \neq 0$, $\;$ equation $(2)$ becomes
$2x - y = 1$ $\;\;$ i.e. $\;$ $y = 2x - 1$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$4x^2 + \left(2x - 1\right)^2 - 2x \left(2x - 1\right) = 7$
i.e. $\;$ $4x^2 + 4x^2 - 4x + 1 - 4x^2 + 2x = 7$
i.e. $\;$ $4x^2 -2x - 6 = 0$
i.e. $\;$ $2x^2 -x - 3 = 0$
i.e. $\;$ $\left(2x - 3\right) \left(x + 1\right) = 0$
i.e. $\;$ $x = \dfrac{3}{2}$ $\;\;$ or $\;\;$ $x = -1$
When $\;$ $x = \dfrac{3}{2}$, $\;$ we have from equation $(3)$, $\;\;$ $y = \left(2 \times \dfrac{3}{2}\right) - 1 = 2$
and when $\;$ $x = -1$, $\;$ we have from equation $(3)$, $\;\;$ $y = \left[2 \times \left(-1\right)\right] - 1 = -3$
Case 2:
When $\;$ $y = 0$, $\;$ equation $(1)$ becomes
$4x^2 = 7$
i.e. $\;$ $x^2 = \dfrac{7}{4}$ $\implies$ $x = \pm \dfrac{\sqrt{7}}{2}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{3}{2}, 2\right), \; \left(-1, -3\right), \; \left(\dfrac{\sqrt{7}}{2}, 0\right), \; \left(\dfrac{- \sqrt{7}}{2}, 0\right) \right\}$