Solve for $x$: $\;\;\;$ $\left|x+2\right| = 2 \left(3-x\right)$
Given problem: $\;\;\;$ $\left|x+2\right| = 2 \left(3-x\right)$ $\;\;\; \cdots \; (1)$
$\left|x+2\right| = x + 2$ $\;$ when $\;$ $x + 2 > 0$ $\;$ i.e. $\;$ $x > -2$
When $\;$ $\left|x+2\right| = x + 2$, $\;$ equation $(1)$ becomes
$x + 2 = 2 \left(3-x\right)$
i.e. $\;$ $x + 2 = 6 - 2x$
i.e. $\;$ $3x = 4$ $\implies$ $x = \dfrac{4}{3}$
$\because \;$ $x > -2$ $\;$ and $\;$ $\dfrac{4}{3} > 2$ $\implies$ $x = \dfrac{4}{3}$ $\;$ is a valid solution.
$\left|x+2\right| = -\left(x + 2\right)$ $\;$ when $\;$ $x + 2 < 0$ $\;$ i.e. $\;$ $x < -2$
When $\;$ $\left|x+2\right| = - \left(x + 2\right)$, $\;$ equation $(1)$ becomes
$-x - 2 = 2 \left(3-x\right)$
i.e. $\;$ $-x - 2 = 6 - 2x$
i.e. $\;$ $x = 8$
$\because \;$ $x < -2$ $\;$ and $\;$ $8 \nless -2$ $\implies$ $x = 8$ $\;$ is not a valid solution.
$\therefore \;$ Solution of equation $(1)$ is $x = \dfrac{4}{3}$