The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$. What is $b - a$?
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$
Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$. 
$a\le 2x+3\le b$
Subtract $3$ from all of the quantities:
$a-3\le 2x\le b-3$
Divide all of the quantities by $2$.
$\frac{a-3}{2}\le x\le \frac{b-3}{2}$
Since we have the range of the solutions, we can make it equal to $10$.
$\frac{b-3}{2}-\frac{a-3}{2} = 10$
Multiply both sides by 2.
$(b-3) - (a-3) = 20$
Re-write without using parentheses.
$b-3-a+3 = 20$
Simplify.
$b-a = 20$
We need to find $b - a$ for the problem, so the answer is $\boxed{20\ \textbf{(D)}}$