Question

Решите систему уравнений
{10^1+lg(x+y)=40,
{lg(x-y)+lg(x+y)=3lg2;

Answer 1

$\left \{ {{10^{1+lg(x+y)}=40} \atop {lg(x-y)+lg(x+y)=3lg2}} \right. \; \left \{ {{10\cdot 10^{lg(x+y)}=40} \atop {lg(x-y)(x+y)=lg2^3}} \right. \; \left \{ {{10^{lg(x+y)}=4} \atop {x^2-y^2=8}} \right. \\\\ \left \{ {{x+y=4} \atop {x^2-y^2=8}} \right.\; \left \{ {{y=4-x} \atop {x^2-(4-x)^2=8}} \right.\\\\x^2-(16-8x+x^2)=8\\\\8x-16=8\\\\8x=24\\\\x=3\\\\y=4-3=1\\\\ODZ:\; \; \left \{ {{x\ \textgreater \ y} \atop {x\ \textgreater \ -y}} \right. \left \{ {{3\ \textgreater \ 1} \atop {3\ \textgreater \ -1}} \right. \\\\Otvet:\; \; (3,1)\; .$