Question

Решить систему, подробнее пожалуйста ​

Answer 1

$\left\{\begin{array}{ccc}9^{x} \cdot 3^{y+1}=729 \\\lg(x+y+4)=1\end{array}\right\\\\ODZ:x+y+4>0\\\\1)9^{x} \cdot 3^{y+1}=729\\\\(3^{2}) ^{x} \cdot 3^{y+1}=3^{6}\\\\3^{2x} \cdot 3^{y+1}=3^{6}\\\\3^{2x+y+1}=3^{6} \\\\2x+y+1=6\\\\\boxed{2x+y=5}\\\\2)\lg(x+y+4)=1\\\\\lg(x+y+4)=\lg10\\\\x+y+4=10\\\\\boxed{x+y=6}\\\\\\-\left\{\begin{array}{ccc}2x+y=5\\x+y=6\end{array}\right\\--------\\x=-1\\\\y=6-x=6-(-1)=6+1=7\\\\Otvet:\boxed{(-1 \ ; \ 7)}$

Answer 2

$\displaystyle\bf ODZ:x+y>-4\\\\ \left \{ {{9^x\cdot3^{y+1}=729} \atop {lg(x+y+4)=1}} \right. \\\\\\\left \{ {{3^{2x}\cdot 3^{y+1}=3^6} \atop {lg(x+y+4)}=lg10} \right. => \\\\\\\left \{ {{3^{2x+y+1}=3^6 \atop {x+y+4=10}} \right. =>\\\\\\\left \{ {{2x+y+1=6} \atop {x+y+4=10}} \right. =>-\left \{ {{2x+y=5} \atop {x+y=6} \right. =>x=-1 \quad ; \quad y=7 \\\\\ ODZ:x+y>-4 => 7-1=6>-4\\\\\ Otvet:\boxed{x=-1 \quad; \quad y=7}$