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

[latex] left { {{2x-y=1} atop {(frac{3^{y}}{27})=(frac{1}{9})^{x-2}}} ight. \ \ left { {{2x-1=y} atop {(frac{3^{y}}{27})=(frac{1}{9})^{x-2}}} ight.[/latex][latex](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}} [/latex][latex]2x-4=4-2x \ 2x+2x=4+4 \ 4x=8 \ x=8/4 \ x=2[/latex][latex]y=2*2-1 \ y=4-1 \ y=3[/latex]