# 2010 AMC 10A Problems/Problem 11

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## 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 1

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

## Solution 2

Without loss of generality, let the interval of solutions be $[0, 10]$ (or any real values $[p, 10+p]$). Then, substitute $0$ and $10$ to $x$. This gives $b=23$ and $a=3$. So, the answer is $23-3=\boxed{20\ \textbf{(D)}}$. ~ bearjere

~IceMatrix