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 $2x-3$ $&lt;$  $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is  presented in interval notation (-$\infty$ , $\frac{3}{2}$].

Any help is appreciated.
To have $|y| = -y$, you need to have $y \leq 0$; now set $y = 2x-3$, so you want $2x-3 \leq 0 \Leftrightarrow x \leq {3 \over 2} \Leftrightarrow x \in (-\infty, {3 \over 2}]$.
case -  when $|2x-3| = -2x+3$ so than there will be $-2x+3=-2x+3$ --- from what will result $0=0$ so for this case the answer will be E).


2.case - when $|2x-3| = 2x-3$ so than there will be $2x-3=-2x+3$ --- 4x=6 --- $x=6/4 =3/2$ --- so in this case $x=3/2$

Note that because in your exercise , in this equation there is sign of equality and not is inequality for this you not can getting interval solutions