Solve the following system of equations: $\;$ $\left|x\right| + 2 \left|y\right| = 3, \;\; 5y + 7x = 2$
Given system of equations: $\;$ $\left|x\right| + 2 \left|y\right| = 3$ $\;\;\; \cdots \; (1)$; $\;$ $5y + 7x = 2$ $\;\;\; \cdots \; (2)$
Case 1:
When $\;$ $x > 0, \; y > 0$, $\;$ $\left|x\right| = + x, \; \left|y\right| = + y$
Then equation $(1)$ becomes $\;\;$ $x + 2y = 3$ $\;\;\; \cdots \; (3)$
Solving equations $(2)$ and $(3)$ simultaneously gives $\;$ $x = \dfrac{-11}{9}, \; y = \dfrac{19}{9}$
But $\;$ $x = \dfrac{-11}{9}$ $\;$ is not a valid solution since it is required that $\;$ $x > 0$.
Case 2:
When $\;$ $x > 0, \; y < 0$, $\;$ $\left|x\right| = + x, \; \left|y\right| = - y$
Then equation $(1)$ becomes $\;\;$ $x - 2y = 3$ $\;\;\; \cdots \; (4)$
Solving equations $(2)$ and $(4)$ simultaneously gives $\;$ $x = 1, \; y = -1$
$\left(x, y\right) = \left(1, -1\right)$ $\;$ is a valid solution since it is required that $\;$ $x > 0, \; y < 0$.
Case 3:
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When $\;$ $x < 0, \; y > 0$, $\;$ $\left|x\right| = - x, \; \left|y\right| = + y$
Then equation $(1)$ becomes $\;\;$ $- x + 2y = 3$ $\;\;\; \cdots \; (5)$
Solving equations $(2)$ and $(5)$ simultaneously gives $\;$ $x = \dfrac{-11}{19}, \; y = \dfrac{23}{19}$
$\left(x, y\right) = \left(\dfrac{-11}{19}, \dfrac{23}{19}\right)$ $\;$ is a valid solution since it is required that $\;$ $x < 0, \; y > 0$.
Case 4:
When $\;$ $x < 0, \; y < 0$, $\;$ $\left|x\right| = - x, \; \left|y\right| = - y$
Then equation $(1)$ becomes $\;\;$ $- x - 2y = 3$ $\;\;\; \cdots \; (6)$
Solving equations $(2)$ and $(6)$ simultaneously gives $\;$ $x = \dfrac{19}{9}, \; y = \dfrac{-23}{9}$
But $\;$ $x = \dfrac{23}{9}$ $\;$ is not a valid solution since it is required that $\;$ $x < 0$.
$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(1, -1\right), \; \left(\dfrac{-11}{19}, \dfrac{23}{19}\right) \right\}$