# 2000 AMC 12 Problems/Problem 5

The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.

## Problem

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

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

## Solution

When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$.

Thus $x-p = (2-p)-p = 2-2p$. $\boxed{\mathbf{(C)}\ \ensuremath{2-2p}}$

## Solution 2 (guess and check/desperation)

If you did not find that slick Solution 1, all hope is not lost. We could still guess and check our way to the right answer.

We first plug in $x=1$, and get that $p=1$ too. Hence $x-p=0$, eliminating choices $A$ and $B$.

We then plug in $x=0$, and get $p=2$. Therefore, $x-p=-2$. The answer is negative, eliminating $E$. Furthermore, $2p-2=2(2)-2=4-2=2\neq-2$, so choice $D$ is false. Hence, the answer must be $C$, which upon checking indeed still holds true. -Monkey_King

## Video Solution by Daily Dose of Math

~Thesmartgreekmathdude