2021 AMC 12B Problems/Problem 3

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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$x=\frac{3}{4} \implies \boxed{\textbf{(A) }$.

Video Solution by OmegaLearn (Algebraic Manipulations)

https://youtu.be/WskJI8_7Gk0

~ pi_is_3.14

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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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