Question

Помогите решить систему уравнений, пожалуйста

Answer 1

Получится вот так :)

Answer 2

$\left\{\!\begin{aligned} & 2^x = \frac{72}{3^y} \\ & 4 \cdot \frac{72}{3^y}+\frac{3^y}{3}=35 \ \cdot | 3 \cdot 3^y \end{aligned}\right. \ \ \ \left\{\!\begin{aligned} & 2^x = \frac{72}{3^y} \\ & 864+3^{2y}-105 \cdot 3^y=0 \end{aligned}\right. \\ \\ 864+3^{2y}-105 \cdot 3^y=0$

Пусть $t=3^y \ (t\ \textgreater \ 0)$
$t^2-105 \cdot t+864=0; \ \ t_{1,2}=\frac{105 \pm \sqrt{11025-4 \cdot 864}}{2}=\frac{105 \pm 87}{2}; \\ \\ t_1=96; \ \ t_2=9; \\ \\ 3^y=96; \ \ 3^y=9; \ \ \ y=\log_3{96}; \ \ y=2$

$\left\{\!\begin{aligned} & 2^x = \frac{72}{3^y} \\ & y=\log_3{96}; \ \ y=2 \end{aligned}\right. \\ \\ \\ 2^x=\frac{72}{3^{\log_3{96}}}=\frac{72}{96}=\frac{3}{4}; \ \ x=\log_2{\frac{3}{4}}; \\ \\ 2^x=\frac{72}{3^2}=\frac{72}{9}=8; \ \ x=3 \\ \\ \\ \left\{\!\begin{aligned} & x=\log_2{\frac{3}{4}, \ \ x=3} \\ & y=\log_3{96}; \ \ y=2 \end{aligned}\right.$