3.1. Solve the system of equations:

$$
\left\{\begin{array}{l}
2^{x+2 y}+2^{x}=3 \cdot 2^{y} \\
2^{2 x+y}+2 \cdot 2^{y}=4 \cdot 2^{x}
\end{array}\right.
$$
Answer: $x=y=\frac{1}{2}$.

Solution.

$$
\begin{gathered}
\left\{\begin{array}{l}
2^{x} \cdot\left(2^{y}\right)^{2}+2^{x}=3 \cdot 2^{y}, \\
\left(2^{x}\right)^{2} \cdot 2^{y}+2 \cdot 2^{y}=4 \cdot 2^{x}
\end{array}\right. \\
2^{x}=u(>0), \quad 2^{y}=v(>0) \\
\left\{\begin{array}{l}
u v^{2}+u-3 v=0, \\
u^{2} v+2 v-4 u=0 .
\end{array}\right. \\
u v^{2}-u^{2} v+u-2 v-3 v+4 u=0, \\
u v(v-u)+5(u-v)=0, \\
u v(v-u)-5(v-u)=0, \\
(u v-5)(v-u)=0, \\
{\left[\begin{array}{c}
u v=5 \\
v-u=0 \quad\left[\begin{array}{l}
u=\frac{5}{v} \\
u=v
\end{array}\right. \\
u=\frac{5}{v} \\
\frac{5}{v} \cdot v^{2}+\frac{5}{v}-3 v=0 \\
5 v+\frac{5}{v}-3 v=0 \\
2 v+\frac{5}{v}=0 \\
2 v^{2}+5=0 \\
(\text { no solutions) }
\end{array}\right.} \\
u=v \\
v^{3}-2 v=0 \\
v=0, \quad v= \pm \sqrt{2} \\
2^{x}=2^{\frac{1}{2}} \\
(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)
\end{gathered}
$$

Answer: $\left(\frac{1}{2}, \frac{1}{2}\right)$