smc

Let's consider each answer choice. $\boxed{A.}\frac{x+y}{y}=\frac{x}{y}+\frac{y}{y}=\frac{3}{4}+1=\boxed{\frac{7}{4}}$ $\boxed{B.}\frac{y-x}{y}=\frac{y}{y}-\frac{x}{y}=1-\frac{3}{4}=\frac{1}{4}⟹{\left(\frac{y-x}{y}\right)}^{-1}=\frac{y}{y-x}=\boxed{\frac{4}{1}}$ $\boxed{C.}\frac{x+2y}{x}=\frac{x}{x}+\frac{2y}{x}=1+2\cdot \frac{4}{3}=\boxed{\frac{11}{3}}$ $\boxed{D.}\frac{x}{2y}=\frac{1}{2}\cdot \frac{3}{4}=\boxed{\frac{3}{8}}$ $\boxed{E.}\frac{x-y}{y}=\frac{x}{y}-\frac{y}{y}=\frac{3}{4}-1=\boxed{-\frac{1}{4}}\ne \frac{1}{4}$ Hence, the incorrect expression is $\boxed{\frac{x-y}{y}=\frac{1}{4}}$. Note: If you're bad at fraction manipulation, you can plug in $x=3$ and $y=4$ and find which expression is false.