2010 AMC 10A Problem 11

Splitting the inequality into two of them and solving gives us $$a \le 2x + 3$$$$x \ge \dfrac{a - 3}{2}$$ and $$2x + 3 \le b$$$$x \le \dfrac{b - 3}{2}.$$ The range of the solutions is then $$\dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10,$$ which then simplifying gives us $$(b - 3) - (a - 3) = 20$$ $$b - a = 20.$$ Thus, D is the correct answer.