Question

Помогите решить систему 2^x+2^y=122^x+y=32

Answer 1

$\left \{ {{2^{x}+2^{y}=12} \atop {2^{x+y}=32}} \right. \left \{ {{2^{x}+2^{y}=12} \atop {2^{x+y}=2^{5}}} \right. \left \{ {{2^{x}+2^{y}=12} \atop {x+y = 5}} \right. \left \{ {{2^{x}+2^{y}=12} \atop {x=5-y}} \right. \\ \\ 2^{5-y} + 2^{y} = 12 (2^{5}/2^{y}) + 2^{y} = 12 | *2^y 2^5 + 2^{2y} = 12*2^y\\\\ 2^y = a\\\\32 + a^2 = 12a\\a^2 - 12a +32 =0\\\\a_1 + a_2 = 12\\a_1 * a_2 = 32\\\\a_1 = 4\\a_2 =8\\\\2^{y_1} = 4\\y_1 = 2\\\\2^{y_2} = 8\\y_2 = 3\\\\x_1 = 5 - y_1 = 5 -2 = 3\\\\x_2 = 5-y_2=5 - 3 = 2$

Ответ: x_1 = 3, y_1 = 2; x_2 = 2, y_2 = 3.