<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] For any two functions $f$ and $g$, we have
$$\begin{align}
|f(x)|-|g(x)|\gt0
&\iff|f(x)|\gt|g(x)|\\
&\iff(f(x))^2\gt(g(x))^2\\
&\iff(f(x))^2-(g(x))^2\gt0\\
&\iff(f(x)-g(x))(f(x)+g(x))\gt0
\end{align}$$
For $f(x)=3x-5$ and $g(x)=2x+3$, we see that $f(x)-g(x)=x-8$ and $f(x)+g(x)=5x-2$, and so
$$|3x-5|-|2x+3|\gt0\iff(x-8)(5x-2)\gt0$$
Since $8\gt2/5$, the two factor are both positive if $x\gt8$ and both negative if $x\lt2/5$, hence
$$|3x-5|-|2x+3|\gt0\iff x\gt8\lor x\lt2/5$$
Remark: This approach is essentially the same as that in Bernard's answer (which appeared while I was composing), the main difference being that the step from $(f(x))^2-(g(x))^2\gt0$ to $(f(x)-g(x))(f(x)+g(x))\gt0$ gives the quadratic in factored form.
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