Solve the following system of equations: $\;$ $x + 2y - 6 = 0, \;\; \left|x - 3\right| - y = 0$
Given system of equations: $\;$ $x + 2y - 6 = 0$ $\;\;\; \cdots \; (1)$
$\left|x - 3\right| - y = 0$ $\;\;\; \cdots \; (2)$
When $\;$ $x - 3 > 0$, $\;$ i.e. $\;$ when $\;$ $x > 3$, $\;$ $\left|x - 3\right| = x - 3$
Then equation $(2)$ becomes
$x - 3 - y = 0$ $\;\;\; \cdots \; (3)$
Solving equations $(1)$ and $(3)$ simultaneously gives
$3x - 12 = 0$ $\implies$ $x = 4$
The condition for this solution is $\;$ $x > 3$
and $\;$ $x = 4 > 3$
$\implies$ $x = 4$ $\;$ is a valid solution.
Substituting the value of $x$ in equation $(1)$ gives
$4 + 2y - 6 = 0$ $\;$ i.e. $\;$ $2y = 2$ $\;$ i.e. $\;$ $y = 1$
When $\;$ $x - 3 < 0$, $\;$ i.e. $\;$ when $\;$ $x < 3$, $\;$ $\left|x - 1\right| = - \left(x - 3\right) = 3 - x$
Then $(2)$ becomes
$3 - x - y = 0$ $\;\;\; \cdots \; (4)$
Solving equations $(1)$ and $(4)$ simultaneously gives
$x = 0$
The condition for this solution is $\;$ $x < 3$
and $\;$ $x = 0 < 3$
$\implies$ $x = 0$ $\;$ is a valid solution.
Substituting the value of $x$ in equation $(1)$ gives
$2y - 6 = 0$ $\implies$ $y = 3$
$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 1\right), \; \left(0, 3\right) \right\}$