Помогите решить систему уравнений

X=3 y=1

$\left \{ {{log_ {\sqrt{2}} (x-y)=2 } \atop {2^x*5^{x-2y}=40 }} \right. \\ \\ log_ {\sqrt{2}} (x-y)=2 \\ x-y= (\sqrt{2} )^2=2 \\ \\ \left \{ {{x-y=2 } \atop {2^x*5^{x-2y}=40 }} \right. \ \\ \left \{ {{x=2 +y} \atop {2^{2+y}*5^{(2+y)-2y}=40 }} \right. \\ \\ 2^{2+y}*5^{(2+y)-2y}=40 \\ 2^{2+y}*5^{2-y}=40\\ 2^2*2^y*5^2* \frac{1}{5^y} =40 \\ 4*2^y*25* \frac{1}{5^y} =40 \\ 100 *\frac{2^y}{5^y} =40 \\ (\frac{2}{5})^y = \frac{40}{100} \\ 0.4^y=0.4 \\ y=1 \\ \\ \left \{ {{x=2+y} \atop {y=1}} \right.$
$\\ \left \{ {{x=2+1} \atop {y=1}} \right. \\ \left \{ {{x=3} \atop {y=1}} \right.$