Difference between revisions of "2006 AMC 10B Problems/Problem 12"
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So <math> a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}</math>. | So <math> a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}</math>. | ||
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+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=q1mxhiyciSk | ||
== See Also == | == See Also == |
Revision as of 23:55, 24 November 2023
Contents
[hide]Problem
The lines and intersect at the point . What is ?
Solution
Since is a solution to both equations, plugging in and will give the values of and .
So .
Video Solution
https://www.youtube.com/watch?v=q1mxhiyciSk
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.