imc

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 $\boxed{20}$