Difference between revisions of "2010 AMC 10B Problems/Problem 13"
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== Solution == | == Solution == | ||
'''Case 1''': | '''Case 1''': | ||
+ | |||
<math> | <math> | ||
2x-|60-2x|=x | 2x-|60-2x|=x | ||
x=|60-2x| | x=|60-2x| | ||
</math> | </math> | ||
− | ''Case 1a'': | + | |
+ | ''Case 1a'': | ||
+ | |||
<math> | <math> | ||
x=60-2x | x=60-2x | ||
Line 26: | Line 29: | ||
x=20 | x=20 | ||
</math> | </math> | ||
+ | |||
''Case 1b'': | ''Case 1b'': | ||
+ | |||
<math> | <math> | ||
-x=60-2x | -x=60-2x | ||
x=60 | x=60 | ||
</math> | </math> | ||
+ | |||
'''Case 2''': | '''Case 2''': | ||
+ | |||
<math> | <math> | ||
2x-|60-2x|=-x | 2x-|60-2x|=-x | ||
3x=|60-2x| | 3x=|60-2x| | ||
</math> | </math> | ||
+ | |||
''Case 2a'': | ''Case 2a'': | ||
+ | |||
+ | <math> | ||
3x=60-2x | 3x=60-2x | ||
5x=60 | 5x=60 | ||
x=12 | x=12 | ||
+ | </math> | ||
+ | |||
+ | ''Case 2b'': | ||
+ | |||
<math> | <math> | ||
− | |||
− | |||
-3x=60-2x | -3x=60-2x | ||
-x=60 | -x=60 | ||
x=-60 | x=-60 | ||
− | <math> | + | </math> |
− | Since an absolute value cannot be negative, we exclude < | + | Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12=92</math> |
Revision as of 20:13, 24 January 2011
Problem
What is the sum of all the solutions of ?
Solution
Case 1:
Case 1a:
Case 1b:
Case 2:
Case 2a:
Case 2b:
Since an absolute value cannot be negative, we exclude . The answer is