4. Solve the system $\left\{\begin{array}{l}3 x \geq 2 y+16 \\ x^{4}+2 x^{2} y^{2}+y^{4}+25-26 x^{2}-26 y^{2}=72 x y .\end{array}\right.$
Answer. $(6 ; 1)$.

Solution. Transform the equation of the system (add $36 x^{2}+36 y^{2}$ to both sides):

$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+25+10 x^{2}+10 y^{2}=36 x^{2}+36 y^{2}+72 x y \Leftrightarrow\left(x^{2}+y^{2}+5\right)^{2}=(6 x+6 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } + 5 = 6 x + 6 y , } \\
{ x ^ { 2 } + y ^ { 2 } + 5 = - 6 x - 6 y }
\end{array} \Leftrightarrow \left[\begin{array}{l}
(x-3)^{2}+(y-3)^{2}=13 \\
(x+3)^{2}+(y+3)^{2}=13
\end{array}\right.\right.
\end{gathered}
$$

We obtain two circles of radius $\sqrt{13}$ with centers at points $(3 ; 3)$ and $(-3 ;-3)$.

The inequality of the system defines a half-plane. Consider the relative position of each circle with the line $y=\frac{3 x}{2}-8$, which is the boundary of this half-plane.

a) $\left\{\begin{array}{l}(x-3)^{2}+(y-3)^{2}=13 \\ y=\frac{3}{2} x-8\end{array} \Leftrightarrow\left\{\begin{array}{l}(x-3)^{2}+\left(\frac{3 x}{2}-11\right)^{2}=13, \\ y=\frac{3}{2} x-8\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{13}{4} x^{2}-39 x+117=0, \\ y=\frac{3}{2} x-8\end{array} \Leftrightarrow\left\{\begin{array}{l}x=6, \\ y=1 .\end{array}\right.\right.\right.\right.$

b) $\left\{\begin{array}{l}(x+3)^{2}+(y+3)^{2}=13, \\ y=\frac{3}{2} x-8\end{array} \Leftrightarrow\left\{\begin{array}{l}(x+3)^{2}+\left(\frac{3 x}{2}-5\right)^{2}=13, \\ y=\frac{3}{2} x-8\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{13}{4} x^{2}-9 x+21=0, \\ y=\frac{3}{2} x-8\end{array} \Leftrightarrow \varnothing\right.\right.\right.$.

In this case, the centers of the considered circles - points $(-3 ;-3)$ and $(3 ; 3)$ - do not lie in the half-plane, as their coordinates do not satisfy the inequality. Therefore, the second circle has no common points with the half-plane, while the first circle has a unique common point $(6 ; 1)$.