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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.

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5 Answers 5

up vote 1 down vote accepted

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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Does that mean $\left(-\infty, \frac{3}{2}\right) \cup {\left \{ \frac{3}{2} \right \}}$ –  FaMu Mar 17 '12 at 21:05
    
@FaMu Yes, it does! : ) –  Matt N. Mar 17 '12 at 21:17
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To have $|y| = -y$, you need to have $y \leq 0$; now set $y = 2x-3$, so you want $2x-3 \leq 0 \Leftrightarrow x \leq {3 \over 2} \Leftrightarrow x \in (-\infty, {3 \over 2}]$.

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To have |y|=−y, you need to have y < 0 (not y ≤ 0); –  FaMu Mar 17 '12 at 20:55
    
@FaMu: if $y=0$, then $-y=0$ and $|y|=0$ and the two are equal... –  Najib Idrissi Mar 18 '12 at 8:47
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$|2x-3|=\begin{cases} 3-2x, & \text{if } x \leq \frac{3}{2} \\ 2x-3, & \text{if } x > \frac{3}{2} \end{cases}$

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] $

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Isn't $\left | 2x-3 \right |= \begin{cases} 2x-3 & \text{ if } x \geq \frac{3}{2} \\ -(2x-3) & \text{ if } x < \frac{3}{2} \end{cases}$ –  FaMu Mar 17 '12 at 20:52
    
@FaMu yes it is...so what ? –  pedja Mar 18 '12 at 8:12
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$$\left | 2x-3 \right | = -(2x-3)$$

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

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

$$t=<0$$

$$2x-3=<0$$

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

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the solution show that $x \in \left(-\infty, \frac{3}{2}\right)$ –  FaMu Mar 17 '12 at 21:00
    
@FaMu thanks for noticing it, I edited my answer –  Tomarinator Mar 18 '12 at 14:31
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  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

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-1: Your answer is incorrect. All of the other answers that have been posted provide the correct answer - take a look at them. Note that one can get an interval even if there is an equality, instead of an inequality: for example, the solutions to the equation $$|t|=t$$ are $t\in[0,\infty)$. –  Zev Chonoles Mar 17 '12 at 9:09
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