2000 AMC 12 Problems/Problem 5: Difference between revisions

Revision as of 12:10, 13 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:

$x-2=-p$

$x=2-p$

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

Revision as of 18:24, 5 November 2006 view source Xantos C. Guin (talk \| contribs) editor 580 edits added category and link to previous and next problem ← Older edit	Revision as of 12:10, 13 November 2006 view source JBL (talk \| contribs) Administrators 2,310 edits m 2000 AMC 12/Problem 5 moved to 2000 AMC 12 Problems/Problem 5 Newer edit →
(No difference)