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.
 A: Whenever you tackle absolute value algebra like this, first find your critical points. Here:
$$3x-5=0\implies x=\frac 53;\ 2x+3=0\implies x=-\frac 32$$
So we have three cases to deal with: $x\geq \frac 53;\ -\frac32\leq x<\frac53;\ x<-\frac32$
In the first, both moduli are positive, so $3x-5>2x+3\implies x>8$, as you got.
In the second, only $|2x+3|$ is positive, so $-(3x-5)>2x+3\implies x<\frac25$
Taking into account our range for the second case, we see a solution here in $-\frac32<x<\frac25$.
Can you deal with the third case, where $x<-\frac32$?
A: Needless to explicit the absolute values in function of the intervals where $x$ lives:
$$|3x-5| > |2x+3| \iff (3x-5)^2>(2x+3)^2\iff 5x^2-42x+16>0,$$
so it comes down to a quadratic inequality.
The reduced discriminant is $\Delta'=21^2-80=361=19^2$, and the quadratic  is positive outside the interval of the roots.
A: 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.
A: Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$ be points on the $xy$-plane. Then the points that divide the line segment $\overline{AB}$ in the ratio $m:n$ are
$$ P = \left( \frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \right), \quad P = \left( \frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n} \right). $$
The point $P$ is the internal divisor and the point $Q$ the external divisor.
We want $x \in \mathbb R$ such that
$$ 3 \left|x-\frac{5}{3}\right| > 2 \left|x-\left(-\frac{3}{2}\right)\right|. $$
Take $A=\left(-\frac{3}{2},0\right)$ and $B=\left(\frac{5}{3},0\right)$, and look for points $P$ and $Q$ that divide the line segment $\overline{AB}$ in the ratio $3:2$.
The formulae above give
$$ P = \left(\frac{(3 \cdot \frac{5}{3})+(2 \cdot -\frac{3}{2})}{3+2},0 \right) = \left(\frac{2}{5},0 \right), \quad Q = \left(\frac{(3 \cdot \frac{5}{3})-(2 \cdot -\frac{3}{2})}{3-2},0 \right) = (8,0). $$
Therefore,
$$ 3 \left|x-\frac{5}{3}\right| > 2 \left|x-\left(-\frac{3}{2}\right)\right| \Longleftrightarrow x>8 \:\:\text{or}\:\: x<\frac{2}{5}. $$
A: $\left|3x-5\right|$ 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
$$x<\frac{2}{5}\quad \mathrm{or}\quad \:x>8$$
