Difference between revisions of "2020 AMC 10A Problems/Problem 1"
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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 21:33, 31 August 2020
Problem
What value of 
satisfies ![\[x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?\]](http://latex.artofproblemsolving.com/5/b/f/5bfe02ab9c82120a6496887cf9a11c3438b9be67.png)
Solution
Adding 
to both sides, 
.
Solution 2
Multiplying 
on both sides gets us 
, therefore 
. ~CoolJupiter
Video Solution
~IceMatrix
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
~savannahsolver
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
