Difference between revisions of "2010 AMC 10A Problems/Problem 11"
Line 19: | Line 19: | ||
<math> (b-3) - (a-3) = 20 </math> | <math> (b-3) - (a-3) = 20 </math> | ||
− | + | Re-write without using parentheses. | |
− | <math> b-3-a+3 = 20 | + | <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> | We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math> |
Revision as of 18:14, 16 August 2010
Since we are given the range of the solutions, we must re-write the inequalities so that we have in terms of
and
.
Subtract from all of the quantities:
Divide all of the quantities by .
Since we have the range of the solutions, we can make them equal to .
Multiply both sides by 2.
Re-write without using parentheses.
Simplify.
We need to find for the problem, so the answer is