Solve the following system of equations: $\;$ $x^2 + 2xy + y^2 - x - y = 6, \;\; x - 2y = 3$
Given system of equations:
$x^2 + 2xy + y^2 - x - y = 6$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $x - 2y = 3$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$,
$x = 2y + 3$ $\;\;\; \cdots \; (2a)$
Substituting the value of $x$ from equation $(2a)$ in equation $(1)$ gives
$\left(2y + 3\right)^2 + 2 \left(2y + 3\right) y + y^2 - \left(2y + 3\right) - y = 6$
i.e. $\;$ $4y^2 + 12y + 9 + 4y^2 + 6y + y^2 - 2y - 3 - y = 6$
i.e. $\;$ $9y^2 + 15y = 0$
i.e. $\;$ $3y \left(3y + 5\right) = 0$
i.e. $\;$ $y = 0$ $\;\;$ or $\;\;$ $y = \dfrac{-5}{3}$
Substituting $y = 0$ in equation $(2a)$ gives $\;\;$ $x = 3$
Substituting $y = \dfrac{-5}{3}$ in equation $(2a)$ gives $\;\;$ $x = \dfrac{-10}{3} + 3 = \dfrac{-1}{3}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(3, 0\right), \left(\dfrac{-1}{3}, \dfrac{-5}{3}\right) \right\}$