1.

\begin{array}{l}{\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}} \\ {\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}}\end{array}

Answer»

thanks guys



Discussion

No Comment Found