Difference between revisions of "Proportion/Introductory"
(includeonly) |
(category) |
||
(2 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | < | + | </noinclude> |
Suppose <math>\frac{1}{20}</math> is either '''x''' or '''y''' in the following system: | Suppose <math>\frac{1}{20}</math> is either '''x''' or '''y''' in the following system: | ||
<cmath>\begin{cases} | <cmath>\begin{cases} | ||
Line 7: | Line 7: | ||
\end{cases} </cmath> | \end{cases} </cmath> | ||
Find the possible values of '''k'''. | Find the possible values of '''k'''. | ||
− | |||
==Solution== | ==Solution== | ||
If <math>x=\frac{1}{20}</math>, then <br /> | If <math>x=\frac{1}{20}</math>, then <br /> | ||
Line 24: | Line 23: | ||
:<math>k=\pm 20</math><br /> | :<math>k=\pm 20</math><br /> | ||
Thus, the possible values of '''k''' are <math>(20,-20)</math>. | Thus, the possible values of '''k''' are <math>(20,-20)</math>. | ||
+ | |||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 13:01, 23 November 2007
Problem
Suppose is either x or y in the following system:
Find the possible values of k.
Solution
If , then
and
Solving gets us:
Thus, there is no solution when
If , then
Thus, the possible values of k are .