Difference between revisions of "2010 AMC 10B Problems/Problem 13"
(→Solution) |
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Line 50: | Line 50: | ||
3x=60-2x, | 3x=60-2x, | ||
5x=60, | 5x=60, | ||
− | x=12 | + | x=12 |
</math> | </math> | ||
Line 58: | Line 58: | ||
-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= \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 20:18, 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