Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 3"

Latest revision as of 20:36, 13 March 2017

Find $x$ and $y$ if $\frac{1}{1+\frac{1}{x}}=2$ and $\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{y}}}}=2$

Solve for each variable.

$\frac{1}{1+\frac{1}{x}}=2$ $1 = 2 + \frac{2}{x}$ $-1 = \frac{2}{x}$ $\boxed{x=-2}$

Now for $y$ :

$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{y}}}}=2$

$1 = 2 + \frac{2}{1+\frac{1}{1+\frac{1}{y}}}$

$-1 = + \frac{2}{1+\frac{1}{1+\frac{1}{y}}}$

$-1 - \frac{1}{1+\frac{1}{y}} = 2$

$\frac{1}{1+\frac{1}{y}} = -3$

$-4 = \frac{3}{y}$

$\boxed{y = -\frac{3}{4}}$

~IYN~

@@ Line 5: / Line 5: @@
 == Solution ==
+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 ==
-{{UNC Math Contest box|year=2014|n=II|num-b=2|num-a=4}}
+{{UNCO Math Contest box|year=2014|n=II|num-b=2|num-a=4}}
 [[Category:Introductory Algebra Problems]]

2014 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All UNCO Math Contest Problems and Solutions