Difference between revisions of "1968 AHSME Problems/Problem 14"
m (→Solution) |
(→Solution) |
||
| Line 10: | Line 10: | ||
== Solution == | == 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 == | == See also == | ||
Revision as of 09:14, 29 May 2023
Problem
If 
and 
are non-zero numbers such that 
and 
, then equals
Solution
We see after multiplying the first equation by , that
Similarly, we see that after multiplying the second equation by , we get that
Thus 
, giving us our final answer of 
~SirAppel
See also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
