Solve the following system of equations: $\;$ $u + 2v = 2, \;\; \left|2u - 3v\right| = 1$
Given system of equations: $\;$ $u + 2v = 2$ $\;\;\; \cdots \; (1)$
$\left|2u - 3v\right| = 1$ $\;\;\; \cdots \; (2)$
When $\;$ $2u - 3v > 0$, $\;$ i.e. $\;$ when $\;$ $2u > 3v$, $\;$ $\left|2u - 3v\right| = 2u - 3v$
Then equation $(2)$ becomes
$2u - 3v = 1$ $\;\;\; \cdots \; (3)$
Solving equations $(1)$ and $(3)$ simultaneously gives
$-7v = -3$ $\implies$ $v = \dfrac{3}{7}$
Substituting $v = \dfrac{3}{7}$ in equation $(1)$ gives
$u + \dfrac{6}{7} = 2$ $\;$ i.e. $\;$ $u = 2 - \dfrac{6}{7} = \dfrac{8}{7}$
Now, when $u = \dfrac{8}{7}$ and $v = \dfrac{3}{7}$,
$2u = \dfrac{16}{7}$, $3v = \dfrac{9}{7}$ and the condition $2u > 3v$ is satisfied.
When $\;$ $2u - 3v < 0$, $\;$ i.e. $\;$ when $\;$ $2u < 3v$, $\;$ $\left|2u - 3v\right| = -2u + 3v$
Then equation $(2)$ becomes
$-2u + 3v = 1$ $\;\;\; \cdots \; (4)$
Solving equations $(1)$ and $(4)$ simultaneously gives
$7v = 5$ $\implies$ $v = \dfrac{5}{7}$
Substituting $v = \dfrac{5}{7}$ in equation $(1)$ gives
$u + \dfrac{10}{7} = 2$ $\;$ i.e. $\;$ $u = 2 - \dfrac{10}{7} = \dfrac{4}{7}$
Now, when $u = \dfrac{4}{7}$ and $v = \dfrac{5}{7}$,
$2u = \dfrac{8}{7}$, $3v = \dfrac{15}{7}$ and the condition $2u < 3v$ is satisfied.
$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(u, v\right) = \left\{\left(\dfrac{8}{7}, \dfrac{3}{7}\right), \; \left(\dfrac{4}{7}, \dfrac{5}{7}\right) \right\}$