Question

Решите систему уравнений:
2x-y=1
(3^y/27)=(1/9)^x-2

Answer 1

$left { {{2x-y=1} atop {(frac{3^{y}}{27})=(frac{1}{9})^{x-2}}} right. \ \ left { {{2x-1=y} atop {(frac{3^{y}}{27})=(frac{1}{9})^{x-2}}} right.$

$(frac{3^{2x-1}}{27})=(frac{1}{9})^{x-2}} \ \ (3^{2x-1}:27)=(frac{1}{3})^{2(x-2)}} \ \ (frac{3^{2x}}{3^1}*frac{1}{27})=(frac{1}{3})^{2x-4}} \ \ (frac{3^{2x}}{3^1}*frac{1}{3^3})=(frac{1}{3})^{2x-4}} \ \ (frac{3^{2x}}{3^4})=(frac{1}{3})^{2x-4}} \ \ 3^{2x-4}=(3^{-1})^{2x-4}} \ \ 3^{2x-4}=3^{4-2x}}$

$2x-4=4-2x \ 2x+2x=4+4 \ 4x=8 \ x=8/4 \ x=2$

$y=2*2-1 \ y=4-1 \ y=3$