7.286. $\left\{\begin{array}{l}\lg \sqrt{(x+y)^{2}}=1, \\ \lg y-\lg |x|=\lg 2 .\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\lg \sqrt{(x+y)^{2}}=1, \\
\lg y-\lg |x|=\lg 2 .
\end{array}\right.
\]
## Solution.

Domain of definition: $\left\{\begin{array}{l}y>0 \\ x \neq 0 \\ x+y \neq 0 .\end{array}\right.$

Write the system of equations as $\left\{\begin{array}{l}\lg |x+y|=1, \\ \lg \frac{y}{|x|}=\lg 2\end{array} \Leftrightarrow\left\{\begin{array}{l}|x+y|=10, \\ \frac{y}{|x|}=2 .\end{array}\right.\right.$

This system of equations is equivalent to four systems:

1) $\left\{\begin{array}{l}x-x, \\ x+y=10, \\ y=-2 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x-x, \\ x=-10, \\ y=-2 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x-x, \\ x=-10, \\ y=20\end{array} \Leftrightarrow\left\{\begin{array}{l}x=-10, \\ y=20 ;\end{array}\right.\right.\right.\right.$
3) $\left\{\begin{array}{l}x>0, \\ y>-x, \\ x+y=10 \\ y=2 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x>0, \\ y>-x, \\ x=\frac{10}{3} \\ y=\frac{20}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{10}{3} \\ y=\frac{20}{3}\end{array}\right.\right.\right.$
4) $\left\{\begin{array}{l}x>0, \\ y0, \\ y<-x, \\ x=-\frac{10}{3}, \text { does not fit. } \\ y=-\frac{20}{3}\end{array}\right.\right.$

Answer: $(-10 ; 20),\left(\frac{10}{3} ; \frac{20}{3}\right)$.