Problem 10.2. A real number $a$ is such that of the two equations

$$
5+|x-2|=a \text { and } 7-|2 x+6|=a
$$

one has exactly one solution, and the other has exactly two solutions. What can $a$ be? List all possible options.
Answer: 5, 7.

Solution. Consider the first equation, equivalent to the equation $|x-2|=a-5$. If $a-5>0$, then it has exactly two roots $x=$ $2 \pm(a-5)$. If $a-5=0$, i.e., $a=5$, then it has exactly one root $x=2$.

Now consider the second equation, equivalent to the equation $|2 x+6|=7-a$. If $7-a>0$, then it has exactly two roots $x=$ $-3 \pm \frac{7-a}{2}$. If $7-a=0$, i.e., $a=7$, then it has exactly one root $x=-3$.

We have found that when $a=5$, the first equation has one root 2, and the second equation has two roots -4 and -2, so such $a$ fits; and when $a=7$, the second equation has one root -3, and the first equation has roots 0 and 4, so such $a$ also fits.