<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] In your second case you write $2x - 3 < 0$. I don't understand how you get $\implies 0 = 0$.
From $2x - 3 < 0$ you get $2x < 3$ and hence $x < \frac{3}{2}$.
Now you take the union of your two sets of solutions to get $x \leq \frac{3}{2}$, or in other words, $x \in (-\infty , \frac{3}{2}]$
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