Прошу, помогите решить

$\left \{ {{4^x\cdot2^y=32} \atop {2^{2x}-2^y=14}} \right.$

$4^x\cdot2^y=32\\ 2^{2x}= \frac{32}{2^y}\\\\ \frac{32}{2^y}-2^y=14\\ 2^y=t\\\\ \frac{32}{t}-t=14\\\\ \frac{32-t^2}{t}=14\\\\ 32-t^2=14t\\ -t^2-14t+32=0\\ t^2+14t-32=0\\ D=196+128=324; \ \sqrt{D}=18\\\\ t_{1/2}= \frac{-14\pm18}{2}\\\\ t_1=2; \ t_2=-16$

Обратная замена:

$2^y=2\\ 2^y=2^1\\ y=1\\\\ 2^y=16$
Не существует

$4^x\cdot2^1=32\\ 4^x= \frac{32}{2}\\ 4^x=16\\ 4^x=4^2\\ x=2$

Итак, $x=2, \ y=1$

Проверка:

$\left \{ {{4^x\cdot2^y=32} \atop {2^{2x}-2^y=14}} \right.\Rightarrow \left \{ {{4^2\cdot2^1=32} \atop {2^{2\cdot2}-2^1=14}} \right. \Rightarrow \left \{ {{16\cdot2=32} \atop {16-2=14}} \right. \Rightarrow \left \{ {{32=32} \atop {14=14}} \right.$

Ответ: $x=2; \ y=1$