2010 AMC 10A Problem 11

Get $x$ by itself, then measure the interval.

1. Start with $a \le 2x + 3 \le b$. Subtract $3$ from all three parts:
$$a - 3 \le 2x \le b - 3.$$
2. Divide all three parts by $2$:
$$\frac{a - 3}{2} \le x \le \frac{b - 3}{2}.$$
3. The length of an interval is the right end minus the left end:
$$\frac{b - 3}{2} - \frac{a - 3}{2} = \frac{(b - 3) - (a - 3)}{2} = \frac{b - a}{2}.$$
4. This length is $10$, so $\dfrac{b - a}{2} = 10$ and $b - a = 20$.

The answer is $\boxed{\textbf{(D)}\ 20}$.

Adding or subtracting the same number from every part of an inequality just slides the interval along; its length stays the same. Dividing every part by $2$ shrinks the interval to half its length. So the interval for $x$ is half as long as the interval from $a$ to $b$.

Try an example. Pick an easy interval for $x$ of length $10$, like $0 \le x \le 10$. Then $2x + 3$ runs from $2(0) + 3 = 3$ to $2(10) + 3 = 23$. So $a = 3$, $b = 23$, and $b - a = 20$.

Isolate $x$ in the compound inequality. Subtracting $3$ throughout and then dividing by $2$ gives
$$\frac{a - 3}{2} \le x \le \frac{b - 3}{2}.$$

The solution interval runs from $\dfrac{a-3}{2}$ to $\dfrac{b-3}{2}$, so its length is
$$\frac{b - 3}{2} - \frac{a - 3}{2} = \frac{b - a}{2}.$$

Setting this equal to $10$ gives $b - a = 20$.

The answer is $\boxed{\textbf{(D)}\ 20}$.