47.2. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=80 \\
\lg (x+y)=\lg x+\lg y
\end{array}\right.
$$

$$
\text { (9-10 grades) }
$$
47.2. Given the condition $x>0, y>0$, the given system is equivalent to the following:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=80 \\
x y=x+y
\end{array}\right.
$$

Multiplying both sides of the second equation by 2 and adding it to the first equation, we arrive at the following equation:

$$
(x+y)^{2}-2(x+y)-80=0
$$

Since $x+y>0$, from the obtained equation we will have:

$$
x+y=1+\sqrt{1+80}=1+9=10
$$

34

Thus, we arrive at the system: $\left\{\begin{array}{l}x+y=10, \\ x y=10,\end{array}\right.$ where $x>0, y>0$. Finally, we get:

$$
\begin{array}{ll}
x_{1}=5+\sqrt{15}, & y_{1}=5-\sqrt{15} \\
x_{2}=5-\sqrt{15}, & y_{2}=5+\sqrt{15}
\end{array}
$$