What value of x satisfies x-\frac{3}{4}=\frac{5}{12}-\frac{1}{3}?
-\frac{2}{3}
\frac{7}{36}
\frac{7}{12}
\frac{2}{3}
\frac{5}{6}
Adding \frac{3}{4} to both sides, x=\frac{5}{12}-\frac{1}{3}+\frac{3}{4}=\frac{5}{12}-\frac{4}{12}+\frac{9}{12}=(\text{E}) \frac{5}{6}.