Solve the following system of equations: $\;$ $\left|x - 1\right| + y = 0, \;\; 2x - y = 1$
Given system of equations: $\;$ $\left|x - 1\right| + y = 0$ $\;\;\; \cdots \; (1)$
$2x - y = 1$ $\;\;\; \cdots \; (2)$
When $\;$ $x - 1 > 0$, $\;$ i.e. $\;$ when $\;$ $x > 1$, $\;$ $\left|x - 1\right| = x - 1$
Then $(1)$ becomes
$x - 1 + y = 0$ $\;\;$ i.e. $\;$ $x + y = 1$ $\;\;\; \cdots \; (3)$
Solving equations $(2)$ and $(3)$ simultaneously gives
$3x = 2$ $\implies$ $x = \dfrac{2}{3}$
The condition for this solution is $\;$ $x > 1$
But $\;$ $x = \dfrac{2}{3} < 1$
$\implies$ $x = \dfrac{2}{3}$ $\;$ is not a valid solution.
When $\;$ $x - 1 < 0$, $\;$ i.e. $\;$ when $\;$ $x < 1$, $\;$ $\left|x - 1\right| = - \left(x - 1\right) = 1 - x$
Then $(1)$ becomes
$1 - x + y = 0$ $\;$ i.e. $\;$ $x - y = 1$ $\;\;\; \cdots \; (4)$
Solving equations $(2)$ and $(4)$ simultaneously gives
$x = 0$
The condition for this solution is $\;$ $x < 1$
and $\;$ $x = 0 < 1$
$\implies$ $x = 0$ $\;$ is a valid solution.
Substituting the value of $x$ in equation $(2)$ gives
$-y = 1$ $\implies$ $y = -1$
$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(0, -1\right) \right\}$