2000 AMC 10 Problem 9

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

An absolute value is a piecewise function, and the inequality $x < 2$ is there precisely to tell you which piece applies. Once the bars are removed correctly, the rest is one substitution. Note that $p > 0$ here, so $2 - 2p$ is negative and cannot equal the always-nonnegative $|2p - 2|$.

Test values. With $x = 0$: $p = 2$ and $x - p = -2$; this kills (B), (D) and (E), which give $2$, $2$, $2$. With $x = 1$: $p = 1$ and $x - p = 0$; this kills (A). Only (C) survives both tests.

• Dropping the absolute value as x - 2 = p, which leads to x - p = 2, choice (B).
• Picking (E) on the feeling that the answer "should" contain an absolute value, even though $x - p$ is negative.