\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(3\cdot\left(2x-y\right)=2\cdot\left(x+y\right)\)
\(6x-3y=2x+2y\)
\(6x-3y-2x-2y=0\)
\(4x-5y=0\)
\(4x=5y\)
\(\frac{x}{y}=\frac{5}{4}\)
Ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\rightarrow6x-3y=2x+2y\)
\(\rightarrow6x-2x=2y+3y\)
\(\rightarrow4x=5y\)
\(\rightarrow\frac{x}{y}=\frac{5}{4}\)