Решить систему. Очень нужно, помогите пжл

Решите задачу:

$\left \{ {{9^{x}-2^y}=1} \atop {9^{-x}-2^{-y}=-\frac{1}{6}}} \right. \; \; \left \{ {{9^{x}-2^{y}=1} \atop {\frac{1}{9^{x}}-\frac{1}{2^{y}}=- \frac{1}{6} }} \right. \; \; \left \{ {{9^{x}-2^{y}=1} \atop { \frac{2^{y}-9^{x}}{9^{x}\cdot 2^{y}}=-\frac{1}{6} }} \right. \; \left \{ {{9^{x}-2^{y}=1} \atop {\frac{-1}{9^{x}\cdot 2^{y}}=-\frac{1}{6}}} \right. \\\\ \left \{ {{9^{x}-2^{y}=1} \atop {9^{x}\cdot 2^{y}=6}} \right. \; \; \left \{ {{\frac{6}{2^{y}}-2^{y}=1} \atop {9^{x}=\frac{6}{2^{y}}} \right.$

$t=2^{y}\ \textgreater \ 0\; ,\; \; \frac{6}{t}-t-1=0\; ,\; \; \frac{6-t^2-t}{t}=0\; ,\\\\t^2+t-6=0\; \; \; \to \; \; \; t_1=-3\ \textless \ 0\; ,\; \; t_2=2\\\\ 2^{y}=2\; \; \; \to \; \; \; y=1\\\\9^{x}=\frac{6}{2}\; ,\; \; 9^{x}=3\; \; \to \; \; 3^{2x}=3\; ,\; \; 2x=1\; ,\; x=\frac{1}{2}\\\\Otvet:\; \; (\frac{1}{2};1)\; .$