Solve the following system of equations: $\;$ $y + x - 1 = 0, \; \left|y\right| - x - 1 = 0$
Given system of equations: $\;$ $y + x - 1 = 0$ $\;\;\; \cdots \; (1)$
$\left|y\right| - x - 1 = 0$ $\;\;\; \cdots \; (2)$
When $\;$ $y > 0$, $\;$ $\left|y\right| = + y$
Then $(2)$ becomes
$y - x - 1 = 0$ $\;\;\; \cdots \; (3)$
Adding equations $(1)$ and $(3)$ gives
$2y - 2 = 0$ $\implies$ $y = 1$
Substituting $\;$ $y = 1$ $\;$ in equation $(1)$ gives
$x = 1 - 1 = 0$
When $\;$ $y < 0$, $\;$ $\left|y\right| = -y$
Then $(2)$ becomes
$-y - x - 1 = 0$
i.e. $\;$ $x + y = -1$ $\;\;\; \cdots \; (4)$
From equation $(1)$, $\;$ $x + y = 1$ $\;\;\; \cdots \; (1a)$
$\therefore \;$ When $\;$ $\left|y\right| = - y$, $\;$ we see from equations $(1a)$ and $(4)$ that equations $(1)$ and $(4)$ cannot be solved simultaneously.
$\therefore \;$ The required solution is $\;$ $\left(x, y\right) = \left\{\left(0, 1\right) \right\}$