Difference between revisions of "2020 AMC 10A Problems/Problem 1"
Aditinair13 (talk | contribs) (→Video Solution 4) |
m (→Solution 2) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 9: | Line 9: | ||
==Solution 2== | ==Solution 2== | ||
− | Multiplying <math>12</math> on both sides gets us <math>12x-9=1</math>, therefore <math>\boxed{x=\textbf{(E)}~\frac{5}{6}}</math>. | + | Multiplying <math>12</math> on both sides gets us <math>12x-9=1 \Rightarrow 12x=10</math>, therefore <math>\boxed{x=\textbf{(E)}~\frac{5}{6}}</math>. |
==Video Solution 1== | ==Video Solution 1== |
Latest revision as of 16:04, 31 March 2021
Contents
Problem
What value of satisfies
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.