Proportion/Introductory: Difference between revisions

Revision as of 16:51, 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 $(20,-20)$ .

@@ Line 1: / Line 1: @@
+<noinclude>
 ==Problem==
-<includeonly>
+</noinclude>
 Suppose <math>\frac{1}{20}</math> is either '''x''' or '''y''' in the following system:
 <cmath>\begin{cases}
@@ Line 7: / Line 8: @@
 \end{cases} </cmath>
 Find the possible values of '''k'''.
-</includeonly>
+<noinclude>
 ==Solution==
 If <math>x=\frac{1}{20}</math>, then <br />
@@ Line 24: / Line 25: @@
 :<math>k=\pm 20</math><br />
 Thus, the possible values of '''k''' are <math>(20,-20)</math>.
+</noinclude>