Solve the following system of equations: $\;$ $xy + 3y^2 - x + 4y - 7 = 0, \;\; 2xy + y^2 - 2x - 2y + 1 = 0$
Given system of equations:
$xy + 3y^2 - x + 4y - 7 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and
$2xy + y^2 - 2x - 2y + 1 = 0$ $\;\;\; \cdots \; (2)$
Multiply equation $(1)$ with $2$ to get
$2xy + 6y^2 - 2x + 8y - 14 = 0$ $\;\;\; \cdots \; (3)$
Subtracting equation $(2)$ from equation $(3)$ gives
$5y^2 + 10y - 15 = 0$
i.e. $\;$ $y^2 + 2y - 3 = 0$
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 $(1)$
$-3x + 27 -x - 12 - 7 = 0$
i.e. $\;$ $4x = 8$ $\implies$ $x = 2$
When $\;$ $y = +1$, $\;$ we have from equation $(1)$
$x + 3 - x + 4 - 7 = 0$
i.e. $\;$ $0 = 0$
i.e. $\;$ $y = +1$ $\;$ is satisfied for all values of $\;$ $x \in R$.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, -3\right), \; \left(p, +1\right) \mid p \in R \right\}$