Difference between revisions of "2006 AMC 12B Problems/Problem 8"
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<math>4x-4a=\frac{1}{4}x+b</math> | <math>4x-4a=\frac{1}{4}x+b</math> | ||
| − | <math>4 | + | <math>4\cdot1-4a=\frac{1}{4}\cdot1+b=2</math> |
<math>a=\frac{1}{2}</math> | <math>a=\frac{1}{2}</math> | ||
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<math>a+b=\frac{9}{4} \Rightarrow \text{(E)}</math> | <math>a+b=\frac{9}{4} \Rightarrow \text{(E)}</math> | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=7|num-a=9}} | {{AMC12 box|year=2006|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:49, 15 September 2016
Problem
The lines 
and 
intersect at the point 
. What is 
?
Solution
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
