Difference between revisions of "2010 AMC 10B Problems/Problem 13"
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== Solution == | == Solution == | ||
− | '''Case 1''': <math>2x-|60-2x|=x</math> | + | '''Case 1''': |
+ | <math> | ||
+ | 2x-|60-2x|=x | ||
+ | x=|60-2x| | ||
+ | </math> | ||
+ | ''Case 1a'': | ||
+ | <math> | ||
+ | x=60-2x | ||
+ | 3x=60 | ||
+ | x=20 | ||
+ | </math> | ||
+ | ''Case 1b'': | ||
+ | <math> | ||
+ | -x=60-2x | ||
+ | x=60 | ||
+ | </math> | ||
+ | '''Case 2''': | ||
+ | <math> | ||
+ | 2x-|60-2x|=-x | ||
+ | 3x=|60-2x| | ||
+ | </math> | ||
+ | ''Case 2a'': | ||
+ | 3x=60-2x | ||
+ | 5x=60 | ||
+ | x=12 | ||
+ | <math> | ||
+ | ''Case 2b'': | ||
+ | </math> | ||
+ | -3x=60-2x | ||
+ | -x=60 | ||
+ | x=-60 | ||
+ | <math> | ||
+ | |||
+ | Since an absolute value cannot be negative, we exclude </math>x=-60<math>. The answer is </math>20+60+12=92$ |
Revision as of 20:12, 24 January 2011
Problem
What is the sum of all the solutions of ?
Solution
Case 1: Case 1a: Case 1b: Case 2: Case 2a: 3x=60-2x 5x=60 x=12 -3x=60-2x -x=60 x=-60 x=-6020+60+12=92$