tubastih7z7h

2023-02-21

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

Jaime Lutz

Beginner2023-02-22Added 9 answers

Locate the zeros of each expression in the modules.

$2x+1=0\Rightarrow x\equiv -\frac{1}{2}$

$3x-2=0\Rightarrow x\equiv \frac{2}{3}$

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:

$x\in ]-\infty ,\frac{1}{2}]\cup [\frac{1}{2},\frac{1}{5}[\cup ]3,+\infty [$

$x\in ]-\infty ,\frac{1}{5}[\cup ]3,+\infty [$

$2x+1=0\Rightarrow x\equiv -\frac{1}{2}$

$3x-2=0\Rightarrow x\equiv \frac{2}{3}$

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:

$x\in ]-\infty ,\frac{1}{2}]\cup [\frac{1}{2},\frac{1}{5}[\cup ]3,+\infty [$

$x\in ]-\infty ,\frac{1}{5}[\cup ]3,+\infty [$