Difference between revisions of "2020 AMC 10A Problems/Problem 1"

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== Solution ==  
 
== Solution ==  
  
Adding <math>\frac{3}{4}</math> to both sides, <math>x= \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12}=\boxed{\textbf{(E) }\frac{5}{6}}</math>.
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==Solution 2==
 
==Solution 2==

Revision as of 18:21, 26 January 2021

Problem

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

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Solution 2

Multiplying $12$ on both sides gets us $12x-9=1$, therefore $\boxed{x=\textbf{(E)}~\frac{5}{6}}$.

Video Solution 1

Education, The Study of Everything

https://youtu.be/4lsmGWDYusk

Video Solution 2

~IceMatrix https://youtu.be/WUcbVNy2uv0

Video Solution 3

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

Video Solution 4

https://youtu.be/OKoBg15l8ro

~savannahsolve

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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