2020 AMC 10A Problem 1

The right side has only numbers in it, so simplify it first. Then undo the "minus $\frac34$".

1. Write both fractions on the right in twelfths. Since $\frac13 = \frac{4}{12}$,
$$\frac{5}{12} - \frac{1}{3} = \frac{5}{12} - \frac{4}{12} = \frac{1}{12}.$$
2. Now the equation reads $x - \frac34 = \frac{1}{12}$.
3. Something minus $\frac34$ gives $\frac{1}{12}$, so add $\frac34$ back to both sides. In twelfths, $\frac34 = \frac{9}{12}$:
$$x = \frac{1}{12} + \frac{9}{12} = \frac{10}{12} = \frac{5}{6}.$$
4. Check: $\frac56 - \frac34 = \frac{10}{12} - \frac{9}{12} = \frac{1}{12}$. It matches.

The answer is $\boxed{\textbf{(E)}\ \frac{5}{6}}$.

To get $x$ alone, you undo what was done to it: it had $\frac34$ taken away, so you add $\frac34$ back. Picking one denominator ($12$) for every fraction at the start turns the problem into adding and subtracting small whole numbers of twelfths.

Work out the right-hand side with denominator $12$:
$$\frac{5}{12} - \frac{1}{3} = \frac{5}{12} - \frac{4}{12} = \frac{1}{12}.$$
So $x - \frac{3}{4} = \frac{1}{12}$. Add $\frac{3}{4} = \frac{9}{12}$ to both sides:
$$x = \frac{1}{12} + \frac{9}{12} = \frac{10}{12} = \frac{5}{6}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{5}{6}}$.