Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 3"
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== Solution == | == Solution == | ||
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+ | Solve for each variable. | ||
+ | |||
+ | <cmath>\frac{1}{1+\frac{1}{x}}=2</cmath> <cmath>1 = 2 + \frac{2}{x}</cmath> <cmath>-1 = \frac{2}{x}</cmath> <cmath>\boxed{x=-2}</cmath> | ||
+ | |||
+ | Now for <math>y</math>: | ||
+ | |||
+ | <cmath>\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{y}}}}=2</cmath> | ||
+ | |||
+ | <cmath>1 = 2 + \frac{2}{1+\frac{1}{1+\frac{1}{y}}}</cmath> | ||
+ | |||
+ | <cmath>-1 = + \frac{2}{1+\frac{1}{1+\frac{1}{y}}}</cmath> | ||
+ | |||
+ | <cmath>-1 - \frac{1}{1+\frac{1}{y}} = 2</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{1+\frac{1}{y}} = -3</cmath> | ||
+ | |||
+ | <cmath>-4 = \frac{3}{y}</cmath> | ||
+ | |||
+ | <cmath>\boxed{y = -\frac{3}{4}}</cmath> | ||
+ | |||
+ | ~IYN~ | ||
== See also == | == See also == | ||
− | {{ | + | {{UNCO Math Contest box|year=2014|n=II|num-b=2|num-a=4}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |