2000 AMC 10 Problem 9

1. Try a number first. Take $x = 1$, which is less than $2$. Then $p = |1 - 2| = |-1| = 1$, not $-1$. The bars turned a negative number positive.
2. In general, $x < 2$ means $x - 2$ is negative. The absolute value of a negative number is its opposite, so
$$p = |x - 2| = -(x - 2) = 2 - x.$$
3. Solve for $x$: add $x$ and subtract $p$ from both sides to get $x = 2 - p$.
4. Now substitute:
$$x - p = (2 - p) - p = 2 - 2p.$$
5. Check with $x = 1$, $p = 1$: $x - p = 0$ and $2 - 2p = 0$. It matches.

The answer is $\boxed{\textbf{(C)}\ 2-2p}$.

Absolute value means "make it positive." The fact $x < 2$ is given so you know $x - 2$ is negative and must be flipped. Once the bars are removed the right way, it is just one substitution. Also, $p$ is positive here, so $2 - 2p$ is negative, and it cannot equal $|2p - 2|$, which is never negative.

Plug in two numbers and cross out choices.
- $x = 0$: then $p = 2$ and $x - p = -2$. Choices (B), (D) and (E) all give $2$, so they are out.
- $x = 1$: then $p = 1$ and $x - p = 0$. Choice (A) gives $-2$, so it is out.

Only (C) is left.

The condition $x < 2$ tells us the sign of $x - 2$: it is negative. The absolute value of a negative number is its opposite, so
$$|x - 2| = -(x - 2) = 2 - x.$$
Hence $p = 2 - x$, which rearranges to $x = 2 - p$. Then
$$x - p = (2 - p) - p = 2 - 2p.$$

The answer is $\boxed{\textbf{(C)}\ 2-2p}$.