# 2010 AMC 10A Problems/Problem 11

## Problem 11

The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$. What is $b - a$? $\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$

## Solution

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

~IceMatrix

## See also

 2010 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2010 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS