Question

Решите систему уравнений:
$\frac{3}{2x+y} + \frac{1}{2x-y} =\frac{2}{5} \\\frac{7}{2x+y} + \frac{2}{2x-y} =\frac{3}{5}$

Answer 1

$\left \{ {{}\frac{3}{2x+y}+ \frac{1}{2x-y}= \frac{2}{5}  \atop {\frac{7}{2x+y}+ \frac{2}{2x-y}= \frac{3}{5}}} \right. \\ \\ 2x+y\neq 0;y\neq -2x;\\ \\ 2x-y\neq 0;y\neq2x \\ \\ \left \{ {{\frac{-6}{2x+y}- \frac{2}{2x-y}= \frac{-4}{5}} \atop {\frac{7}{2x+y}+ \frac{2}{2x-y}= \frac{3}{5}}} \right. \\ \\ ----------\\ \frac{-6+7}{2x+y} =\frac{-1}{5} \\ \\ \frac{1}{2x+y} =-\frac{1}{5} \\ \\ 2x+y=-5\\ \\ y=-2x-5\\ \\ \frac{3}{2x-2x-5}+ \frac{1}{2x+2x+5}= \frac{2}{5}\\ \\ \frac{1}{4x+5} =\frac{2}{5} +\frac{3}{5}$

$\frac{1}{4x+5} =1\\ \\ 4x+5=1\\ \\ 4x=-4\\ \\ x=-1\\ \\ y=-2*(-1)-5=-3\\ \\ OTVET:(-1;-3)$