Difference between revisions of "2010 AMC 10B Problems/Problem 13"
| Line 18: | Line 18: | ||
<math> | <math> | ||
| − | 2x-|60-2x|=x | + | 2x-|60-2x|=x, |
| − | |||
x=|60-2x| | x=|60-2x| | ||
| Line 27: | Line 26: | ||
<math> | <math> | ||
| − | x=60-2x | + | x=60-2x, |
| − | + | 3x=60, | |
| − | 3x=60 | ||
| − | |||
x=20 | x=20 | ||
</math> | </math> | ||
| Line 37: | Line 34: | ||
<math> | <math> | ||
| − | -x=60-2x | + | -x=60-2x, |
x=60 | x=60 | ||
</math> | </math> | ||
| Line 44: | Line 41: | ||
<math> | <math> | ||
| − | 2x-|60-2x|=-x | + | 2x-|60-2x|=-x, |
| − | |||
3x=|60-2x| | 3x=|60-2x| | ||
</math> | </math> | ||
| Line 52: | Line 48: | ||
<math> | <math> | ||
| − | 3x=60-2x | + | 3x=60-2x, |
| − | + | 5x=60, | |
| − | 5x=60 | + | x=12, |
| − | |||
| − | x=12 | ||
</math> | </math> | ||
| Line 62: | Line 56: | ||
<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 <math>x=-60</math>. The answer is <math>20+60+12=92</math> | 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 20:14, 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 
