Difference between revisions of "2021 AMC 12B Problems/Problem 3"
Olympushero (talk | contribs) |
Olympushero (talk | contribs) |
||
Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
Subtracting <math>2</math> from both sides and taking reciprocals gives <math>1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}</math>. Subtracting <math>1</math> from both sides and taking reciprocals again gives <math>2+\frac{2}{3+x}=\frac{38}{15}</math>. Subtracting <math>2</math> from both sides and taking reciprocals for the final time gives <math>\frac{x+3}{2}=\frac{15}{8}</math> or <math>x=\frac{3}{4} \implies \boxed{\text{A}}</math>. | Subtracting <math>2</math> from both sides and taking reciprocals gives <math>1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}</math>. Subtracting <math>1</math> from both sides and taking reciprocals again gives <math>2+\frac{2}{3+x}=\frac{38}{15}</math>. Subtracting <math>2</math> from both sides and taking reciprocals for the final time gives <math>\frac{x+3}{2}=\frac{15}{8}</math> or <math>x=\frac{3}{4} \implies \boxed{\text{A}}</math>. | ||
+ | |||
+ | ~ OlympusHero | ||
== Video Solution by OmegaLearn (Algebraic Manipulations) == | == Video Solution by OmegaLearn (Algebraic Manipulations) == |
Revision as of 22:49, 11 February 2021
Problem
SupposeWhat is the value of
Solution 1
Subtracting from both sides and taking reciprocals gives
. Subtracting
from both sides and taking reciprocals again gives
. Subtracting
from both sides and taking reciprocals for the final time gives
or
.
~ OlympusHero
Video Solution by OmegaLearn (Algebraic Manipulations)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.