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

(Solution 2)
(Solution 2)
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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>.
 
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>.
  
=Solution 2=
+
==Solution 2==
Multiplying <math>12</math> on both sides gets us <math>12x-9=1</math>, therefore <math>x=\frac{5}{6}</math>. ~CoolJupiter
+
Multiplying <math>12</math> on both sides gets us <math>12x-9=1</math>, therefore <math>\boxed{x=\textbf{(C)}\frac{5}{6}}</math>. ~CoolJupiter
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:33, 31 August 2020

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

Adding $\frac{3}{4}$ to both sides, $x= \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12}=\boxed{\textbf{(E) }\frac{5}{6}}$.

Solution 2

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

Video Solution

https://youtu.be/WUcbVNy2uv0

~IceMatrix


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

~bobthefam


https://youtu.be/OKoBg15l8ro

~savannahsolver

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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