The solution set of the equation $|2x - 3| = - (2x - 3)$

Question

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.

Answer 1

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, $x \in (-\infty , \frac{3}{2}]$

Answer 2

To have $|y| = -y$, you need to have $y \leq 0$; now set $y = 2x-3$, so you want $2x-3 \leq 0 \Leftrightarrow x \leq {3 \over 2} \Leftrightarrow x \in (-\infty, {3 \over 2}]$.

Answer 3

1. 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

Answer 4

$|2x-3|=\begin{cases} 3-2x, & \text{if } x \leq \frac{3}{2} \\ 2x-3, & \text{if } x > \frac{3}{2} \end{cases}$

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] $

Answer 5

$$\left | 2x-3 \right | = -(2x-3)$$

$let$, $t= 2x-3$

$$\left | t \right | = -t$$

$$t=<0$$

$$2x-3=<0$$

$$x \in \left(-\infty, \frac{3}{2}\right]$$