Решите систему уравнений : {log√2(x-y)=2; {3^6-x×4^y+3=36

[ $\left \{ {{{log_ {\sqrt{2}} (x-y)=2} \atop {{3^{6-x}\cdot4^{y+3}=36}} \right. \\ \\ \left \{ {{{ (\sqrt{2}) ^{2} = x-y; x-y\ \textgreater \ 0} \atop {{3^{6}\cdot 3^{-x}}\cdot4^{y}\cdot 4^3}=4\cdot 9}} \right. \\ \\ \left \{ {{{ (y = x-2; x-y\ \textgreater \ 0} \atop {{3^{4}\cdot 3^{-x}\cdot4^{y}\cdot 4^{2}}=1}} \right. \\ \\ \left \{ {{{ (y = x-2; x-y\ \textgreater \ 0} \atop {{3^{4-x}\cdot4^{y+2}=1}} \right.$
Что-то не так
Из второго уравнения скорее всего 4-х=0, у+2=0 тогда х=4, у=-2
но тогда х-у= 4-(-2) не равно 2

или

$\left \{ {{{ y = x-2; x-y\ \textgreater \ 0} \atop {{ 3^{6-x}}\cdot4^{x-2+3}}=36}} \right. \\ \\ \left \{ {{{ y = x-2; x-y\ \textgreater \ 0} \atop {{ 3^{6}\cdot 3^{-x}}\cdot4^{x}\cdot4}=36}}} \right. \\ \\\left \{ {{{ y = x-2; x-y\ \textgreater \ 0} \atop {{ 3^{6}\cdot 3^{-x}}\cdot4^{x}\cdot4}=4\cdot 9}}} \right. \\ \\$
$\left \{ {{{ y = x-2; x-y\ \textgreater \ 0} \atop {{( \frac{4}{3}) ^{x}= \frac{1}{81} }} \right. \\ \\$

$x=log_{ \frac{4}{3}} \frac{1}{81} \\ \\ y=log_{ \frac{4}{3}} \frac{1}{81}-2$