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 | + | Multiplying <math>12</math> on both sides gets us <math>12x-9=1</math>, therefore <math>\boxed{x=\textbf{(E)}~\frac{5}{6}}</math>. |
| − | ==Video Solution== | + | ==Video Solution 1== |
| − | https://youtu.be/ | + | Education, The Study of Everything |
| + | |||
| + | https://youtu.be/4lsmGWDYusk | ||
| + | ==Video Solution 2== | ||
~IceMatrix | ~IceMatrix | ||
| + | https://youtu.be/WUcbVNy2uv0 | ||
| − | + | ==Video Solution 3== | |
https://www.youtube.com/watch?v=7-3sl1pSojc | https://www.youtube.com/watch?v=7-3sl1pSojc | ||
~bobthefam | ~bobthefam | ||
| − | + | ==Video Solution 4== | |
https://youtu.be/OKoBg15l8ro | https://youtu.be/OKoBg15l8ro | ||
| − | ~ | + | ~savannahsolve |
== See Also == | == See Also == | ||
Revision as of 18:22, 26 January 2021
Contents
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 
.
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix https://youtu.be/WUcbVNy2uv0
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~savannahsolve
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. 
