Difference between revisions of "2010 AMC 10B Problems/Problem 13"
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| Line 18: | Line 18: | ||
<math> | <math> | ||
| − | 2x-|60-2x|=x | + | 2x-|60-2x|=x |
| + | </math> | ||
| + | |||
| + | <math> | ||
x=|60-2x| | x=|60-2x| | ||
</math> | </math> | ||
| Line 25: | Line 28: | ||
<math> | <math> | ||
| − | x=60-2x | + | x=60-2x |
| − | 3x=60 | + | </math> |
| + | |||
| + | <math> | ||
| + | 3x=60 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
x=20 | x=20 | ||
</math> | </math> | ||
| Line 33: | Line 42: | ||
<math> | <math> | ||
| − | -x=60-2x | + | -x=60-2x |
| + | </math> | ||
| + | |||
| + | <math> | ||
x=60 | x=60 | ||
</math> | </math> | ||
| Line 40: | Line 52: | ||
<math> | <math> | ||
| − | 2x-|60-2x|=-x | + | 2x-|60-2x|=-x |
| + | </math> | ||
| + | |||
| + | <math> | ||
3x=|60-2x| | 3x=|60-2x| | ||
</math> | </math> | ||
| Line 47: | Line 62: | ||
<math> | <math> | ||
| − | 3x=60-2x | + | 3x=60-2x |
| − | 5x=60 | + | </math> |
| + | |||
| + | <math> | ||
| + | 5x=60 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
x=12 | x=12 | ||
</math> | </math> | ||
| Line 55: | Line 76: | ||
<math> | <math> | ||
| − | -3x=60-2x | + | -3x=60-2x |
| − | -x=60 | + | </math> |
| + | |||
| + | <math> | ||
| + | -x=60 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
x=-60 | x=-60 | ||
</math> | </math> | ||
Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12= \boxed{\mathrm {(C)}\ 92}</math> | Since an absolute value cannot be negative, we exclude <math>x=-60</math>. The answer is <math>20+60+12= \boxed{\mathrm {(C)}\ 92}</math> | ||
Revision as of 21:23, 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 
