4. Find the set of values of the expression $x-y+1$ under the condition

$$
(x-y)^{2}=2|2 y-x|+x+15
$$
# Problem 4.

Answer: $(-\infty ;-2] \cup[6 ;+\infty)$.

Solution. Let's perform a change of variables: $x-y+1=a, 2 y-x=t$. Then the condition of the problem can be reformulated as follows: "Find the set of values of $a$ under the condition $2|t|+t=a^{2}-4 a-12$."

On the plane of variables $(a, t)$, this condition defines a set consisting of parts of the parabolas $t=\frac{1}{3}\left(a^{2}-4 a-12\right)$ and $t=-\left(a^{2}-4 a-12\right)$ for values of $a \in(-\infty ;-2] \cup[6 ;+\infty)$. Since there are no points of the set in the strip $-2<a<6$, the specified values of $a$ constitute the answer.