Difference between revisions of "User:Temperal/Introductory Proportion"
| Line 5: | Line 5: | ||
x=ky | x=ky | ||
\end{cases} </cmath> | \end{cases} </cmath> | ||
| − | Find the possible values of '''k'''. | + | Find the possible values of '''k''' in terms of x and y. |
==Solution== | ==Solution== | ||
{{incomplete|solution}} | {{incomplete|solution}} | ||
| − | 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>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>, |
| + | then we solve again, we get <math>k=20x</math> and <math>\frac{20} {x}</math> | ||
Revision as of 13:30, 22 September 2007
Problem
Suppose 
is either x or y in the following system:
![\[\begin{cases} xy=\frac{1}{k}\\ x=ky \end{cases}\]](http://latex.artofproblemsolving.com/4/8/f/48f37af559f7c66fe3ccb3fb9b0e6c16b3fec444.png)
Find the possible values of k in terms of x and y.
Solution
Template:Incomplete
If 
, then 
so 
or 
. now we get 
and 
. If 
,
then we solve again, we get 
and 
