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| − | ==Problem==
| + | #REDIRECT [[Proportion/Introductory]] |
| − | Suppose <math>\frac{1}{20}</math> is either '''x''' or '''y''' in the following system:
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| − | <cmath>\begin{cases}
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| − | xy=\frac{1}{k}\\
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| − | x=ky
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| − | \end{cases} </cmath>
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| − | Find the possible values of '''k'''.
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| − | | |
| − | ==Solution==
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| − | If <math>x=\frac{1}{20}</math>, then <br />
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| − | :<math>\frac{1}{20}=ky</math> and
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| − | :<math>\frac{y}{20}=\frac{1}{k}</math><br />
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| − | Solving gets us:<br />
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| − | :<math>y=\frac{20}{k}</math>
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| − | :<math>\frac{1}{20}=k\frac{20}{k}</math>
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| − | :<math>\frac{1}{20}=20</math><br />
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| − | Thus, there is no solution when <math>x=\frac{1}{20}</math><br />
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| − | If <math>y=\frac{1}{20}</math>, then <br />
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| − | :<math>\frac{x}{20}=\frac{1}{k} \Longrightarrow xk=20</math>
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| − | :<math>x=\frac{k}{20}</math>
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| − | :<math>\left(\frac{k}{20}\right)\cdot k=20</math>
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| − | :<math>k^2=400</math>
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| − | :<math>k=\pm 20</math><br />
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| − | Thus, the possible values of '''k''' are <math>(20,-20)</math>.
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