7. As the real parameter $a$ varies, what is the maximum number of solutions for the equation ||$x-1|-4|+x=a$ ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) it can have infinitely many.
7. The answer is (E). Indeed, for every $x \leq-3, x-1 \leq 0$ and thus $|x-1|=1-x$, furthermore $|x-1|-4=-x-3 \geq 0$, and the equation can be simplified as follows:

$$
|| x-1|-4|+x=|-x+1-4|+x=-x-3+x=-3
$$

Therefore, if $a=-3$ the equation has infinitely many solutions.