Question

помогите алгебра !!!!​

Answer 1

$\left\{\begin{array}{ccc}2^{x}\cdot 3^{y}=12\\2^{y}\cdot 3^{x}=18\end{array}\right\; \; \left\{\begin{array}{c}2^{x}\cdot 3^{y}\cdot 2^{y}\cdot 3^{x}=12\cdot 18\\2^{y}\cdot 3^{x}=18\qquad \qquad \end{array}\right\; \; \left\{\begin{array}{ccc}2^{x+y}\cdot 3^{x+y}=216\\2^{y}\cdot 2^{x}=18\end{array}\right\\\\\\\left\{\begin{array}{ccc}6^{x+y}=6^3\; \; \; \; \\2^{y}\cdot 3^{x}=2\cdot 3^2\end{array}\right\; \; \left\{\begin{array}{ccc}x+y=3\qquad \; \\2^{3-x}\cdot 3^{x}=2\cdot 3^2\end{array}\right\;$

$2^{3-x}\cdot 3^{x}=2\cdot 3^2\; |:(2\cdot 3^2)\\\\2^{2-x}\cdot 3^{x-2}=1\; \; ,\; \; \; (\frac{2}{3})^{2-x}=1\; \; ,\; \; \; (\frac{2}{3})^{2-x}=2^0\; \; ,\; \; 2-x=0\; \; ,\\\\\boxed {x=2}$

$y=3-x=3-2=1\; \; ,\; \; \boxed {y=1}\\\\Otvet:\; \; (2,1)\; .$