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

Revision as of 13:45, 22 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

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

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

$xk=20$

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

$k^2=400$

$k=\pm 20$

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

@@ Line 5: / Line 5: @@
 x=ky
 \end{cases} </cmath>
-Find the possible values of '''k''' in terms of '''x''' and '''y'''.
+Find the possible values of '''k'''.
 ==Solution==
-{{incomplete|solution}}
+If <math>x=\frac{1}{20}</math>, then <br />
-If <math>x=\frac{1} {20}</math>, then <math> \frac{y} {20} = \frac{1} {k}</math> so <math>ky=20</math> or <math>ky=\frac{1} {20}</math>. now we get <math>k=\frac{20} {y}</math> and <math>\frac{1} {20y}</math>. If <math>y=\frac{1} {20}</math>,
+:<math>\frac{1}{20}=ky</math> and
-then we solve again, we get <math>k=20x</math> and <math>\frac{20} {x}</math>
+:<math>\frac{y}{20}=\frac{1}{k}</math><br />
+Solving gets us:<br />
+:<math>y=\frac{20}{k}</math>
+:<math>\frac{1}{20}=k\frac{20}{k}</math>
+:<math>\frac{1}{20}=20</math><br />
+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>x=\frac{k}{20}</math>
+:<math>xk=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>.