# 3.2. Solve the system of equations:

$$
\left\{\begin{array}{l}
2^{x+y}-2^{x-y}=4 \\
2^{x+y}-8 \cdot 2^{y-x}=6
\end{array}\right.
$$
Answer: $x=\frac{5}{2}, y=\frac{1}{2}$.

Solution.

$$
\begin{aligned}
& 2^{x+y}=u(>0), \quad 2^{y-x}=v(>0) \\
& \left\{\begin{array}{l}
u-v=4 \\
u-\frac{8}{v}=6
\end{array}\right. \\
& \left\{\begin{array}{l}
u=v+4 \\
v+4-\frac{8}{v}=6
\end{array}\right. \\
& v^{2}+4 v-6 v-8=0 \\
& v^{2}-2 v-8=0 \\
& {\left[\begin{array}{c}
v=4 \\
v=-2 \text { (extraneous root) }
\end{array} \Rightarrow u=4+4=8\right.} \\
& \left\{\begin{array}{l}
2^{x-y}=4 \\
2^{x+y}=8
\end{array}\right. \\
& \left\{\begin{array}{l}
x-y=2 \\
x+y=3
\end{array}\right. \\
& 2 x=5, \quad 2 y=1 \\
& x=\frac{5}{2}, \quad y=\frac{1}{2}
\end{aligned}
$$