Решите систему уравнений: 2x-y=1(3^y/27)=(1/9)^x-2

Решите задачу:

$\left \{ {{2x-y=1} \atop {(\frac{3^{y}}{27})=(\frac{1}{9})^{x-2}}} \right. \\ \\ \left \{ {{2x-1=y} \atop {(\frac{3^{y}}{27})=(\frac{1}{9})^{x-2}}} \right.$

$(\frac{3^{2x-1}}{27})=(\frac{1}{9})^{x-2}} \\ \\ (3^{2x-1}:27)=(\frac{1}{3})^{2(x-2)}} \\ \\ (\frac{3^{2x}}{3^1}*\frac{1}{27})=(\frac{1}{3})^{2x-4}} \\ \\ (\frac{3^{2x}}{3^1}*\frac{1}{3^3})=(\frac{1}{3})^{2x-4}} \\ \\ (\frac{3^{2x}}{3^4})=(\frac{1}{3})^{2x-4}} \\ \\ 3^{2x-4}=(3^{-1})^{2x-4}} \\ \\ 3^{2x-4}=3^{4-2x}}$

$2x-4=4-2x \\ 2x+2x=4+4 \\ 4x=8 \\ x=8/4 \\ x=2$

$y=2*2-1 \\ y=4-1 \\ y=3$