2021 AMC 12B Problems/Problem 3: Difference between revisions
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<math>\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}</math> | <math>\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}</math> | ||
==Solution 1== | |||
Subtracting <math>2</math> from both sides and taking reciprocals gives 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{\textbf{(A) }$. | |||
== Video Solution by OmegaLearn (Algebraic Manipulations) == | == Video Solution by OmegaLearn (Algebraic Manipulations) == | ||
Revision as of 21:48, 11 February 2021
Problem
Suppose![\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]](http://latex.artofproblemsolving.com/9/f/2/9f2bfdf62a315bd20a6a692df961e22f22f7611d.png)
What is the value of 
Solution 1
Subtracting 
from both sides and taking reciprocals gives 1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}
1
2+\frac{2}{3+x}=\frac{38}{15}2
\frac{x+3}{2}=\frac{15}{8}
x=\frac{3}{4} \implies \boxed{\textbf{(A) }$.
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 | |
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