2010 AMC 10A Problem 11

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 $\boxed{\textbf{(D)}\ 20}$.

Adding a constant to every part of an inequality slides the interval without changing its length; multiplying or dividing by a positive constant scales the length by that constant. Here $2x + 3$ stretches $x$ by $2$, so the $x$-interval is half as long as the interval $[a, b]$.

Pick concrete numbers: if $a = 3$ and $b = 23$, then $3 \le 2x + 3 \le 23$ becomes $0 \le x \le 10$, an interval of length exactly $10$, and $b - a = 20$.

• Forgetting to divide by 2 (answering 10) or thinking the shift by 3 changes the length.
• Dividing the wrong way and doubling the length in the wrong direction, giving $b - a = 5$ (not offered) or confusing which quantity is $10$.