<s>[INST] The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\  2x-3   = -(2x-3)$ when $2x-3$ $\geqslant$  $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$  $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is  presented in interval notation (-$\infty$ , $\frac{3}{2}$].
Any help is appreciated.
 [/INST] $|2x-3|=\begin{cases}
 3-2x, & \text{if } x \leq \frac{3}{2} \\
 2x-3, & \text{if } x > \frac{3}{2}
\end{cases}$
a) $|2x-3|=3-2x$ , hence :
$3-2x=-(2x-3)$
$0=0$ , therefore :
$S_a : x \in \left(-\infty, \frac{3}{2}\right]$
b) $|2x-3|=2x-3$ , hence :
$2x-3=-(2x-3)$
$4x=6$
$x=\frac{3}{2}$ , therefore :
$S_b : x \in \emptyset $
Finally :
$S= S_a \cup S_b \Rightarrow S : x \in \left(-\infty, \frac{3}{2}\right]  $
</s>