tubastih7z7h

2023-02-21

How to solve $|2x+1|<|3x-2|$?

Jaime Lutz

Locate the zeros of each expression in the modules.
$2x+1=0⇒x\equiv -\frac{1}{2}$
$3x-2=0⇒x\equiv \frac{2}{3}$
The equation must now be divided into three parts:
1) When $x\ge 2/3$, both expression are larger than 0, hence:
$|2x+1|<|3x-2|$
is equivalent to:
$2x+1<3x-2\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x\ge \frac{2}{3}$
$-x<-3\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x\ge \frac{2}{3}$
$x>3\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x\ge \frac{2}{3}$
$x>3$
2)if $-\frac{1}{2}\le x<\frac{2}{3}$:
3x-2 will be <=0, therefore $|3x-2|=-3x+2$:
$2x+1<-3x+2\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}-\frac{1}{2}\le x\le \frac{2}{3}$:
$5x<1\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}-\frac{1}{2}\le x\le \frac{2}{3}$:
$x<\frac{1}{5}\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}-\frac{1}{2}\le x\le \frac{2}{3}$:
$-\frac{1}{2}\le x<\frac{1}{5}$
3)if $x\le -\frac{1}{2}$
Given that both statements are negative, we should find their inverses.
$-2x-1<-3x+2\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x\le -\frac{1}{2}$
$x<3\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x<-\frac{1}{2}$
$x\le -\frac{1}{2}$
Now the solution is the union of the three solutions:
$x\in \right]-\infty ,\frac{1}{2}\right]\cup \left[\frac{1}{2},\frac{1}{5}\left[\cup \right]3,+\infty \left[$
$x\in \right]-\infty ,\frac{1}{5}\left[\cup \right]3,+\infty \left[$

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