Difference between revisions of "2000 AMC 12 Problems/Problem 5"

Revision as of 14:22, 16 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 $|x - 2| = 2-x$ and $\displaystyle x - 2 = -p$ .

Therefore:

$x=2-p$

$\displaystyle x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)}$

@@ Line 6: / Line 6: @@
 == Solution ==
-When <math>\displaystyle x < 2,</math>, <math>x-2</math> is negative.
+When <math>\displaystyle x < 2,</math>, <math>x-2</math> is negative so <math>|x - 2| = 2-x</math> and <math>\displaystyle x - 2 = -p</math>.
-So <math>\displaystyle x - 2 = -p,</math>.
 Therefore:
-<math>x-2=-p</math>
 <math>x=2-p</math>
-<math>\displaystyle x-p = (2-p)-p = 2-2p \Rightarrow C </math>
+<math>\displaystyle x-p = (2-p)-p = 2-2p \Longrightarrow \mathrm{(C)} </math>
 == See also ==
 * [[2000 AMC 12 Problems]]
-*[[2000 AMC 12/Problem 4|Previous Problem]]
+*[[2000 AMC 12 Problems/Problem 4|Previous Problem]]
-*[[2000 AMC 12/Problem 6|Next problem]]
+*[[2000 AMC 12 Problems/Problem 6|Next problem]]
 [[Category:Introductory Algebra Problems]]