\(4^{x+y}=64\Rightarrow x+y=3\)
\(2^{2x-y}=128\Rightarrow 2x-y=7\)
So we have
\(x+y=3\)
\(2x-y=7\)
Solve for \(x\):
\(3x=10\)
\(\Rightarrow x=\frac{10}{3}\)
Plug \(x\) into one of the equations:
\(\frac{10}{3}+y=3\)
\(\Rightarrow y=-\frac{1}{3}\)
So the answer is
\(x=\frac{10}{3}, y=-\frac{1}{3}\)
