4. Solve the system $\left\{\begin{array}{l}x+3 y+14 \leq 0, \\ x^{4}+2 x^{2} y^{2}+y^{4}+64-20 x^{2}-20 y^{2}=8 x y .\end{array}\right.$
Answer: $(-2, -4)$.

Solution. Transform the equation of the system (add $4 x^{2}+4 y^{2}$ to both sides):

$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+64-16 x^{2}-16 y^{2}=4 x^{2}+4 y^{2}+8 x y \Leftrightarrow\left(x^{2}+y^{2}-8\right)^{2}=(2 x+2 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } - 8 = 2 x + 2 y , } \\
{ x ^ { 2 } + y ^ { 2 } - 8 = - 2 x - 2 y }
\end{array} \Leftrightarrow \left[\begin{array}{l}
(x-1)^{2}+(y-1)^{2}=10 \\
(x+1)^{2}+(y+1)^{2}=10
\end{array}\right.\right.
\end{gathered}
$$

We obtain two circles of radius $\sqrt{10}$ with centers at points $(1, 1)$ and $(-1, -1)$.

The inequality of the system defines a half-plane. Consider the relative position of each circle with the line $x=-3 y-14$, which is the boundary of this half-plane.

a) $\left\{\begin{array}{l}(x-1)^{2}+(y-1)^{2}=10, \\ x=-3 y-14\end{array} \Leftrightarrow\left\{\begin{array}{l}(-3 y-15)^{2}+(y-1)^{2}=10, \\ x=-3 y-14\end{array} \Leftrightarrow\left\{\begin{array}{l}10 y^{2}+88 x+216=0, \\ x=-3 y-14\end{array} \Leftrightarrow \varnothing\right.\right.\right.$

b) $\left\{\begin{array}{l}(x+1)^{2}+(y+1)^{2}=10, \\ x=-3 y-14\end{array} \Leftrightarrow\left\{\begin{array}{l}(-3 y-13)^{2}+(y+1)^{2}=10, \\ x=-3 y-14\end{array} \Leftrightarrow\left\{\begin{array}{l}10 y^{2}+80 y+160=0, \\ x=-3 y-14\end{array} \Leftrightarrow\left\{\begin{array}{l}y=-4, \\ x=-2 .\end{array}\right.\right.\right.\right.$

In this case, the centers of the considered circles - points $(1, 1)$ and $(-1, -1)$ - do not lie in the half-plane, as their coordinates do not satisfy the inequality. Therefore, the first circle has no common points with the half-plane, while the second circle has a unique common point $(-2, -4)$.