Difference between revisions of "1968 AHSME Problems/Problem 14"

Revision as of 10:14, 29 May 2023

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

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

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

1968 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

@@ Line 10: / Line 10: @@
 == Solution ==
-<math>\fbox{E}</math>
+We see after multiplying the first equation by <math>y</math>, that
+<math>xy=y+1.</math>
+Similarly, we see that after multiplying the second equation by <math>x</math>, we get that
+<math>xy=x+1.</math>
+Thus <math>x+1=y+1 \implies x=y</math>, giving us our final answer of <math>\fbox{E}.</math>
+~SirAppel
 == See also ==