Difference between revisions of "Proportion/Introductory"

Revision as of 17:50, 9 October 2007

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: @@
 ==Problem==
+<includeonly>
 Suppose <math>\frac{1}{20}</math> is either '''x''' or '''y''' in the following system:
 <cmath>\begin{cases}
@@ Line 6: / Line 7: @@
 \end{cases} </cmath>
 Find the possible values of '''k'''.
+</includeonly>
 ==Solution==
 If <math>x=\frac{1}{20}</math>, then <br />