2000 AMC 12 Problems/Problem 5: Difference between revisions

Revision as of 18:24, 5 November 2006

If $\displaystyle |x - 2| = p,$ where $\displaystyle x < 2,$ then $\displaystyle x - p =$

$\mathrm{(A) \ -2 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 2-2p } \qquad \mathrm{(D) \ 2p-2 } \qquad \mathrm{(E) \ |2p-2| }$

When $\displaystyle x < 2,$ , $x-2$ is negative.

So $\displaystyle x - 2 = -p,$ .

Therefore:

$x-2=-p$

$x=2-p$

$\displaystyle x-p = (2-p)-p = 2-2p \Rightarrow C$

@@ Line 18: / Line 18: @@
 <math>\displaystyle x-p = (2-p)-p = 2-2p \Rightarrow C </math>
-== See Also ==
+== See also ==
-[[2000 AMC 12 Problems]]
+* [[2000 AMC 12 Problems]]
+*[[2000 AMC 12/Problem 4|Previous Problem]]
+*[[2000 AMC 12/Problem 6|Next problem]]
+[[Category:Introductory Algebra Problems]]