
Problem 8'.1. Find all values of the real parameter $a$ such that the inequalities $|x+1|+|2-x|<a$ and $\frac{5 a-8}{6 x-5 a+5}<-\frac{1}{2}$ are equivalent.


Solution: We begin with the first inequality. $|x+1|+|2-x|\frac{1-a}{2}$, i. e., $\frac{1-a}{2}3$. If $x \in[-1 ; 2]$, then $|x+1|+|2-x|=x+1+2-x=3$ and the above equation has a solution only when $a>3$. Finally, if $x>2$, then $x+1+x-23$. Therefore when $a \leq 3$, the inequality has no solution, and when $a>3$, the solutions form the interval $\left(\frac{1-a}{2} ; \frac{1+a}{2}\right)$.

Let us rewrite the second inequality in the form $\frac{6 x+5 a-11}{6 x-5 a+5}<$ 0. Its solutions form either the interval $\left(\frac{11-5 a}{6} ; \frac{5 a-5}{6}\right)$ or the interval $\left(\frac{5 a-5}{6} ; \frac{11-5 a}{6}\right)$, depending on which of the two numbers $\frac{5 a-5}{6}$ and $\frac{11-5 a}{6}$ is greater.

Therefore the two inequalities are equivalent if

$$
\frac{1-a}{2}=\frac{11-5 a}{6}, \quad \frac{1+a}{2}=\frac{5 a-5}{6}
$$

or

$$
\frac{1-a}{2}=\frac{5 a-5}{6}, \quad \frac{1+a}{2}=\frac{11-5 a}{6}
$$

The first pair is satisfied by $a=4$, the second one by $a=1$. But if $a=1$, the first inequality has no solution, whereas the second one does. Thus the only solution is $a=4$.
