15 ){\begin{aligned}27^{y}\cdot 3^{x} &= 1\\(\frac{1}{2})^{x}\cdot 4^{y} &= 2\end{aligned}}

$$\begin{aligned}&\begin{cases}27^{y}\cdot 3^{x} = 1\\(\frac{1}{2})^{x}\cdot 4^{y} = 2\end{cases}\\&\begin{cases}(3^{3})^{y}\cdot 3^{x} = 1\\(2^{-1})^{x}\cdot (2^{2})^{y} = 2\end{cases}\\&\begin{cases}3^{3y}\cdot 3^{x} = 3^{0}\\2^{-x}\cdot 2^{2y} = 2^{1}\end{cases}\\&\begin{cases}3y+x = 0\\-x+2y = 1\end{cases}\end{aligned}$$ Из первого уравнения выразим x: $$\begin{aligned}&x = -3y\end{aligned}$$ Подставим значение x во второе уравнение: $$\begin{aligned}&-(-3y)+2y = 1\\&3y+2y = 1\\&5y = 1\\&y = \frac{1}{5}\end{aligned}$$ Тогда: $$\begin{aligned}&x = -3\cdot \frac{1}{5}\\&x = -\frac{3}{5}\end{aligned}$$ Ответ: x=-3/5; y=1/5