Пусть (x;y)- решение системы уравнений. Найти x+y

Question 1

Question 2

Решите задачу:

$\left \{ {{10^{1+lg(x+y)}=40} \atop {lg(x-y)+lg(x+y)=3\, lg2}} \right. \; \; ,\; \; \; ODZ:\; \left \{ {{x-y\ \textgreater \ 0} \atop {x+y\ \textgreater \ 0}} \right. \; ,\; \left \{ {{x\ \textgreater \ y} \atop {x\ \textgreater \ -y}} \right. \\\\a)\; \; 10^{1+lg(x+y)}=4\cdot 10\; ,\; \; \; 10^{1+lg(x+y)}=10^{lg4}\cdot 10\; ,\\\\10^{1+lg(x+y)}=10^{lg4+1}\; \; \Rightarrow \quad 1+lg(x+y)=lg4+1\; ,\\\\lg(x+y)=lg4\; \; \Rightarrow \quad \underline {x+y=4}\; \; ,\; \; y=4-x\\\\b)\; \; lg(x-y)+lg(x+y)=2\, lg2\; \; ,\; \; lg(x-y)(x+y)=lg2^3\; ,\\\\(x-y)(x+y)=8\; ,\; \; (x-y)\cdot 4=8\; ,\; \; x-y=2$

$y=x-2\\\\ c)\; \; \left \{ {{y=4-x} \atop {y=x-2}} \right. \; \; \Rightarrow \; \; 4-x=x-2\; \; \Rightarrow \; \; \; 2x=6\; \; ,\; \; x=3\\\\y=4-x=4-3=1\\\\ODZ:\; x\ \textgreater \ y\; \; \to \; \; 3\ \textgreater \ 1\; \; ;\; \; \; x\ \textgreater \ -y\; \; \to \; \; 3\ \textgreater \ -1\\\\d)\; \; x+y=3+1=4\\\\Otvet:\; \; (3,1)\; \; ,\; \; x+y=4\; .$

$P.S.\; \; \; 10^{1+lg(x+y)}=40\; \; \to \; \; 10\cdot 10^{lg(x+y)}=40\; \; \to \; \; 10^{lg(x+y)}=4\; ,\\\\x+y=4\\\\Formyla:\; \; a^{log_{a}b}=b\; \; \to \; \; 10^{lgb}=b\; ,\; \; b\ \textgreater \ 0\; ,\; a\ \textgreater \ 0\; ,\; a\ne 1\; .$