# 2020 AMC 10A Problems/Problem 1

## Problem 1

What value of $x$ satisfies $$x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?$$

$\textbf{(A)}\ -\frac{2}{3}\qquad\textbf{(B)}\ \frac{7}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{5}{6}$

## Solution

Adding $\frac{3}{4}$ to both sides, $x= \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12}=$\boxed{\text{(E) }\frac{5}{5}}\$.

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