How can I solve inequalities like $|3x-5|<x+2$?

Question

This is a problem from an advanced mathematics exam book.
I'm a bit confused on how to do it because there's an $x$ outside the absolute sign.
I watched a couple of tutorials online and tried solving the problem. Here's what I did:
$3x-5<x+2$

$3x<x+7$

$2x<7$

$x<7/2$
Is my approach correct?

Answer 1

You may not simply disregard the absolute value sign the way you did. Recall that for all $y$, $|y|=y$ if $y\geq 0$ and $|y|=-y$ if $y<0$. Therefore, you have to get rid of the absolute value sign by considering cases.

Case one: $x\geq \frac{5}{3}$. In this case, $3x-5\geq 0$, hence, the inequality becomes

$$ 3x-5<x+2\iff2x<7\iff x<\frac{7}{2}. $$

Hence, the inequality is satisfied if $\frac{5}{3}\leq x<\frac{7}{2}$.

Case two: $x<\frac{5}{3}$. In this case, $3x-5<0$, hence, the inequality becomes

$$ -(3x-5)<x+2\iff-3x+5<x+2\iff 3<4x\iff\frac{3}{4}<x. $$

Thus, the inequality is satisfied if $\frac{3}{4}<x<\frac{5}{3}$.

Combining these two results, you obtain that the inequality is satisfied for all $\frac{3}{4}<x<\frac{7}{2}$.

Answer 2

Note that $\lvert u \lvert\lt a \iff -a\lt u\lt a$ where $a\gt 0$. Hence $$ \begin{align} \lvert 3x-5\lvert\lt x+2&\iff -x-2\lt 3x-5 \lt x+2\\ &\iff -x+3\lt 3x\lt x+7.\\ \end{align}$$ Now solve the two separate inequalities $3x\lt x+7$ and $-x+3\lt 3x$ along with $x+2\gt0$ and take the intersection of their solution sets.

If you solve the inequalities you get $x\lt 7/2,\; x\gt3/4$ and $x\gt-2$ respectively. Together we see that the solution is $3/4\lt x\lt 7/2$.