Difference between revisions of "Proportion/Introductory"

Revision as of 16:52, 9 October 2007

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.

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

$\frac{x}{20}=\frac{1}{k} \Longrightarrow xk=20$

$x=\frac{k}{20}$

$\left(\frac{k}{20}\right)\cdot k=20$

$k^2=400$

$k=\pm 20$

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

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-<noinclude>
 ==Problem==
 </noinclude>
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 \end{cases} </cmath>
 Find the possible values of '''k'''.
-<noinclude>
 ==Solution==
 If <math>x=\frac{1}{20}</math>, then <br />
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 :<math>k=\pm 20</math><br />
 Thus, the possible values of '''k''' are <math>(20,-20)</math>.
-</noinclude>