Question

9 класс,Система рациональных уравнений.
Буду благодарен за помощь!

Answer 1

$\left \{ {{\frac{1}{x+y} +\frac{1}{x-y}=10|*(-3) } \atop {\frac{3}{x+y}+\frac{6}{x-y} =54 }} \right.\\\\+\left \{ {{\frac{-3}{x+y}+\frac{-3}{x-y} =-30 } \atop {\frac{3}{x+y} +\frac{6}{x-y}=54 }} \right. \\ ----------\\\frac{3}{x-y}=24\\\\x-y=\frac{1}{8}\\\\x=y+\frac{1}{8}\\\\\frac{1}{y+\frac{1}{8}+ y}+\frac{1}{\frac{1}{8} }=10\\\\\frac{1}{2y+\frac{1}{8} }+8=10\\\\\frac{1}{2y+\frac{1}{8} }=2\\\\2y+\frac{1}{8}=\frac{1}{2}\\\\2y=\frac{3}{8}\\\\y=\frac{3}{16}$

$x=\frac{3}{16}+\frac{1}{8}=\frac{5}{16}$

Ответ :

$x=\frac{5}{16};y=\frac{3}{16}$

Answer 2

$\left \{ {{\frac{1}{x+y}+\frac{1}{x-y}=10} \atop {\frac{3}{x+y}+\frac{6}{x-y}=54}} \right. \quad t=\frac{1}{x+y}\; \; ,\; \; p=\frac{1}{x-y}\; \; \left \{ {{t+p=10} \atop {3t+6p=54}} \right. \\\\\\\left \{ {{t=10-p} \atop {3(10-p)+6p=54}} \right. \; \left \{ {{t=10-p} \atop {3p=24}} \right. \; \left \{ {{t=2} \atop {p=8}} \right. \\\\\frac{1}{x+y}=2\; \; \to \; \; x+y=\frac{1}{2}\\\\\frac{1}{x-y}=8\; \; \to \; \; x-y=\frac{1}{8}\quad \left \{ {{x+y=\frac{1}{2}} \atop {x-y=\frac{1}{8}}} \right.\\\\2x=\frac{5}{8}\; \; ,\; \; 2y=\frac{3}{8}\\\\x=\frac{5}{16}\; \; ,\; \; y=\frac{3}{16}$

$Otvet:\; \; (\frac{5}{16},\frac{3}{16})\; .$