Difference between revisions of "2010 AMC 10A Problems/Problem 11"

Revision as of 17:14, 16 August 2010

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

@@ Line 19: / Line 19: @@
 <math> (b-3) - (a-3) = 20 </math>
-Simplify
+Re-write without using parentheses.
-<math> b-3-a+3 = 20 \RightArrow b-a = 20 </math>
+<math> b-3-a+3 = 20 </math>
+Simplify.
+<math> b-a = 20 </math>
 We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math>