Solve the following system of equations: $\;$ $3x^2 + xy - 2x + y - 5 = 0, \;\; 2x^2 - xy -3x - y - 5 = 0$
Given system of equations:
$3x^2 + xy - 2x + y - 5 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $2x^2 - xy -3x - y - 5 = 0$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ gives
$5x^2 - 5x - 10 = 0$
i.e. $\;$ $x^2 - x - 2 = 0$
i.e. $\;$ $\left(x - 2\right) \left(x + 1\right) = 0$
i.e. $\;$ $x = 2$ $\;\;$ or $\;\;$ $x = -1$
Substituting $\;$ $x = 2$ $\;$ in equation $(1)$ gives
$12 + 2y - 4 + y - 5 = 0$
i.e. $\;$$3y = -3$ $\implies$ $y = -1$
Substituting $\;$ $x = -1$ $\;$ in equation $(1)$ gives
$3 - y + 2 + y - 5 = 0$
i.e. $\;$ $0 = 0$
$\implies$ $x = -1$ is satisfied for all $y \in R$.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, -1\right), \; \left(-1, p\right) \mid p \in R \right\}$