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Difference between revisions of "2021 AMC 12B Problems/Problem 3"
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==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>.
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~ OlympusHero
  
 
== Video Solution by OmegaLearn (Algebraic Manipulations) ==
 
== Video Solution by OmegaLearn (Algebraic Manipulations) ==

Revision as of 22:49, 11 February 2021

Problem

Suppose\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]What is the value of $x?$

$\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}$

Solution 1

Subtracting $2$ from both sides and taking reciprocals gives $1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}$. Subtracting $1$ from both sides and taking reciprocals again gives $2+\frac{2}{3+x}=\frac{38}{15}$. Subtracting $2$ from both sides and taking reciprocals for the final time gives $\frac{x+3}{2}=\frac{15}{8}$ or $x=\frac{3}{4} \implies \boxed{\text{A}}$.

~ OlympusHero

Video Solution by OmegaLearn (Algebraic Manipulations)

https://youtu.be/WskJI8_7Gk0

~ 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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