Difference between revisions of "User:Temperal/Introductory Proportion"

Revision as of 10:03, 23 September 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 $(20,-20)$ .

@@ Line 17: / Line 17: @@
 Thus, there is no solution when <math>x=\frac{1}{20}</math><br />
 If <math>y=\frac{1}{20}</math>, then <br />
-:<math>\frac{x}{20}=\frac{1}{k}</math>
+:<math>\frac{x}{20}=\frac{1}{k} \Longrightarrow xk=20</math>
 :<math>x=\frac{k}{20}</math>
-:<math>xk=20</math>
+:<math>\left(\frac{k}{20}\right)\cdot k=20</math>
-:<math>\frac{k}{20}\cdot k=20</math>
 :<math>k^2=400</math>
 :<math>k=\pm 20</math><br />
 Thus, the possible values of '''k''' are <math>(20,-20)</math>.