User:Temperal/Introductory Proportion

< User:Temperal
Revision as of 12:45, 22 September 2007 by Temperal (talk | contribs)

Problem

Suppose $\frac{1}{20}$ is either x or y in the following system: \[\begin{cases} xy=\frac{1}{k}\\ x=ky \end{cases}\] Find the possible values of k.

Solution

If $x=\frac{1}{20}$, then

$\frac{1}{20}=ky$ and
$\frac{y}{20}=\frac{1}{k}$

Solving gets us:

$y=\frac{20}{k}$
$\frac{1}{20}=k\frac{20}{k}$
$\frac{1}{20}=20$

Thus, there is no solution when $x=\frac{1}{20}$
If $y=\frac{1}{20}$, then

$\frac{x}{20}=\frac{1}{k}$
$x=\frac{k}{20}$
$xk=20$
$\frac{k}{20}\cdot k=20$
$k^2=400$
$k=\pm 20$

Thus, the possible values of k are $(20,-20)$.