Difference between revisions of "2006 AMC 10B Problems/Problem 12"
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== Problem == | == Problem == | ||
+ | The lines <math> x=\frac{1}{4}y+a </math> and <math> y=\frac{1}{4}x+b </math> intersect at the point <math> (1,2) </math>. What is <math> a+b </math>? | ||
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+ | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4} </math> | ||
+ | |||
== Solution == | == Solution == | ||
+ | Since <math>(1,2)</math> is a solution to both equations, plugging in <math>x=1</math> and <math>y=2</math> will give the values of <math>a</math> and <math>b</math>. | ||
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+ | <math> 1 = \frac{1}{4} \cdot 2 + a </math> | ||
+ | |||
+ | <math> a = \frac{1}{2} </math> | ||
+ | |||
+ | <math> 2 = \frac{1}{4} \cdot 1 + b </math> | ||
+ | |||
+ | <math> b = \frac{7}{4} </math> | ||
+ | |||
+ | So: <math> a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow E </math> | ||
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== See Also == | == See Also == | ||
− | *[[ | + | {{AMC10 box|year=2006|ab=B|num-b=11|num-a=13}} |
+ | |||
+ | * [[Line | Lines]] | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:17, 4 July 2013
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:
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 |
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