13 ){\begin{aligned}(\sqrt{3})^{x+2y} &= \sqrt{3}\cdot \sqrt{27}\\0,1^{x}\cdot 10^{3y} &= 10\end{aligned}}

$$\begin{aligned}&\begin{cases}(\sqrt{3})^{x+2y} = \sqrt{3}\cdot \sqrt{27}\\0,1^{x}\cdot 10^{3y} = 10\end{cases}\\&\begin{cases}(3^{\frac{1}{2}})^{x+2y} = 3^{\frac{1}{2}}\cdot 3^{\frac{3}{2}}\\(\frac{1}{10})^{x}\cdot 10^{3y} = 10\end{cases}\\&\begin{cases}3^{\frac{1}{2}(x+2y)} = 3^{\frac{1}{2} + \frac{3}{2}}\\10^{-x}\cdot 10^{3y} = 10^{1}\end{cases}\\&\begin{cases}\frac{1}{2}(x+2y) = 2\\-x+3y = 1\end{cases}\\&\begin{cases}x+2y = 4\\-x+3y = 1\end{cases}\end{aligned}$$ Cложим уравнения: $$\begin{aligned}&x+2y-x+3y = 4+1\\&5y = 5\\&y = 1\end{aligned}$$ Подставим значение y в первое уравнение: $$\begin{aligned}&x+2\cdot 1 = 4\\&x = 4-2\\&x = 2\end{aligned}$$ Ответ: x=2; y=1