Solve the following system of equations: $\;$ $x + 3 \left|y\right| - 1 = 0, \;\; x + y + 3 = 0$
Given system of equations: $\;$ $x + 3 \left|y\right| - 1 = 0$ $\;\;\; \cdots \; (1)$
$x + y + 3 = 0$ $\;\;\; \cdots \; (2)$
When $\;$ $y > 0$, $\;$ $\left|y\right| = + y$
Then $(1)$ becomes
$x + 3y - 1 = 0$ $\;\;\; \cdots \; (3)$
Solving equations $(2)$ and $(3)$ simultaneously gives
$2y - 4 = 0$ $\implies$ $y = 2$
Substituting $\;$ $y = 2$ $\;$ in equation $(2)$ gives
$x + 2 + 3 = 0$ $\implies$ $x = -5$
When $\;$ $y < 0$, $\;$ $\left|y\right| = -y$
Then $(1)$ becomes
$x - 3y - 1 = 0$ $\;\;\; \cdots \; (4)$
Solving equations $(2)$ and $(4)$ simultaneously gives
$-4y - 4 = 0$ $\implies$ $y = -1$
Substituting the value of $y$ in equation $(2)$ gives
$x - 1 + 3 = 0$ $\implies$ $x = -2$
$\therefore \;$ The required solution is $\;$ $\left(x, y\right) = \left\{\left(-5, 2\right), \; \left(-2, -1\right) \right\}$