Difference between revisions of "1968 AHSME Problems/Problem 14"
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\text{(E) } x</math> | \text{(E) } x</math> | ||
| − | == Solution == | + | == Solution 1 == |
We see after multiplying the first equation by <math>y</math>, that | We see after multiplying the first equation by <math>y</math>, that | ||
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~SirAppel | ~SirAppel | ||
| + | |||
| + | == Solution 2 == | ||
| + | We can first combine the fraction on the second equation, which is | ||
| + | |||
| + | <math>y=\dfrac{x+1}{x}.</math> | ||
| + | |||
| + | Then, substituting the value of <math>x</math> from the first equation, we can get | ||
| + | |||
| + | <math>y=\dfrac{1+\dfrac{1}{y}+1}{1+\dfrac{1}{y}}.</math> | ||
| + | |||
| + | Then, multiplying the denominator on both sides, we get | ||
| + | |||
| + | <math>y+1=2+\dfrac{1}{y} \implies y=1+\dfrac{1}{y}</math> which is equivalent to the value of <math>x</math> that was originally provided, giving the final answer of <math>\fbox{E}.</math> | ||
| + | |||
| + | ~TurtleGod7 | ||
== See also == | == See also == | ||
Latest revision as of 03:01, 10 July 2025
Contents
Problem
If 
and 
are non-zero numbers such that 
and 
, then equals
Solution 1
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
Solution 2
We can first combine the fraction on the second equation, which is
Then, substituting the value of from the first equation, we can get
Then, multiplying the denominator on both sides, we get
which is equivalent to the value of that was originally provided, giving the final answer of
~TurtleGod7
See also
| 1968 AHSC (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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
