2024 AMC 10A Problems/Problem 11

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Problem

If $x$ and $y$ are real numbers satisfying $x + \tfrac{x}{y} = 2$ and $y + \tfrac{y}{x} = 6$, what is the value of $x + y$?

$\textbf{(A)}~\frac{101}{28} \qquad\textbf{(B)}~\frac{42}{11} \qquad\textbf{(C)}~\frac{30}{7} \qquad\textbf{(D)}~\frac{14}{3} \qquad\textbf{(E)}~\frac{110}{21}$

Solution

The given conditions imply that $\tfrac{xy + x}{y} = 2$ and $\tfrac{xy + y}{x} = 6$. Therefore, $xy + x = 2y$ and $xy + y = 6x$. Subtracting the two equations gives $x - y = 2y - 6x$, so $3y = 7x$ and $\tfrac{x}{y} = \tfrac{3}{7}$. Then \[x + \tfrac{3}{7} = 2 \implies x = \tfrac{11}{7}\] and consequently, $y = \tfrac{11}{3}$, so $x + y = \tfrac{11}{7} + \tfrac{11}{3} = \boxed{\textbf{(E)}~\tfrac{110}{21}}$.

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions

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