5.1. Solve the inequality

$$
8 \cdot \frac{|x+1|-|x-7|}{|2 x-3|-|2 x-9|}+3 \cdot \frac{|x+1|+|x-7|}{|2 x-3|+|2 x-9|} \leqslant 8
$$

In the answer, write the sum of its integer solutions that satisfy the condition $|x|<120$.
Answer: 6 (solution of the inequality: $\left.\left[\frac{3}{2} ; 3\right) \cup\left(3 ; \frac{9}{2}\right]\right)$.

Solution. Let $a$ and $b$ be the first and second terms in the left part of the inequality, respectively. Then $b>0$ and

$$
a b=24 \cdot \frac{(x+1)^{2}-(x-7)^{2}}{(2 x-3)^{2}-(2 x-9)^{2}}=24 \cdot \frac{8 \cdot(2 x-6)}{6 \cdot(4 x-12)}=16
$$

for $x \neq 3$, from which $a=\frac{16}{b}$. The inequality $\frac{16}{b}+b \leqslant 8\left(\Leftrightarrow\left(\frac{4}{\sqrt{b}}-\sqrt{b}\right)^{2} \leqslant 0\right)$ is satisfied only when $b=4$. It remains to solve the equation

$$
3 \cdot \frac{|x+1|+|x-7|}{|2 x-3|+|2 x-9|}=4
$$

If $x \geqslant 7$ or $x \leqslant-1$, the expression on the left side equals $\frac{3}{2}$. If $-1<x<\frac{3}{2}$ or $\frac{9}{2}<x<7$, it equals $\frac{6}{|3-x|}<4$. For all $\frac{3}{2} \leqslant x \leqslant \frac{9}{2}$, the equation turns into a true equality. Thus, taking into account the condition $x \neq 3$, we obtain the solution of the original inequality: $\left[\frac{3}{2} ; 3\right) \cup\left(3 ; \frac{9}{2}\right]$. The integer solutions are 2 and 4, both of which satisfy the condition $|x|<120$, their sum is 6.