<s>[INST] Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and $2x-3$ is negative for certain values of $x$, then $$3x-5 > -2x-3$$ implies $$5x >2$$ implies $$x > \dfrac{2}{5}$$ I'm supposed to get that $x < \dfrac{2}{5}$ according to the solutions, but I'm not sure how to get that solution.
 [/INST] $\left|3x-5\right|$ is negative for $x<-\frac{3}{2}$ and $-\frac{3}{2}\le x<\frac{5}{3}$. Positive for $x\ge \frac{5}{3}$.
For $\left|2x+3\right|$ you have in the same intervals the signs: $-$, $+$ and $+$.
Hence for $x<-\frac{3}{2}$ you have $5-3x-\:2x-3\:>\:0 \iff x<\frac{2}{5}$. and so on and so removing the absolute values in relation to the intervals where they are positive or negative.
Combining the intervals you will have:
$$x<-\frac{3}{2}\quad \mathrm{or}\quad \:-\frac{3}{2}\le \:x<\frac{2}{5}\quad \mathrm{or}\quad \:x>8$$ and merge overlapping the intervals
$$x<\frac{2}{5}\quad \mathrm{or}\quad \:x>8$$
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