<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] 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,
$$ 3 \left|x-\frac{5}{3}\right| > 2 \left|x-\left(-\frac{3}{2}\right)\right| \Longleftrightarrow x>8 \:\:\text{or}\:\: x<\frac{2}{5}. $$
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