1968 AHSME Problems/Problem 14

Problem

If $x$ and $y$ are non-zero numbers such that $x=1+\frac{1}{y}$ and $y=1+\frac{1}{x}$, then $y$ equals

$\text{(A) } x-1\quad \text{(B) } 1-x\quad \text{(C) } 1+x\quad \text{(D) } -x\quad \text{(E) } x$

Solution 1

We see after multiplying the first equation by $y$, that

$xy=y+1.$

Similarly, we see that after multiplying the second equation by $x$, we get that

$xy=x+1.$

Thus $x+1=y+1 \implies x=y$, giving us our final answer of $\fbox{E}.$

~SirAppel

Solution 2

We can first combine the fraction on the second equation, which is

$y=\dfrac{x+1}{x}.$

Then, substituting the value of $x$ from the first equation, we can get

$y=\dfrac{1+\dfrac{1}{y}+1}{1+\dfrac{1}{y}}.$

Then, multiplying the denominator on both sides, we get

$y+1=2+\dfrac{1}{y} \implies y=1+\dfrac{1}{y}$ which is equivalent to the value of $x$ that was originally provided, giving the final answer of $\fbox{E}.$

~TurtleGod7

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AHSME Problems and Solutions

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