Difference between revisions of "2006 AMC 10B Problems/Problem 12"

 
 
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== Problem ==
 
== Problem ==
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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>
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== Solution ==
 
== Solution ==
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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>
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<math> a = \frac{1}{2} </math>
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<math> 2 = \frac{1}{4} \cdot 1 + b </math>
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<math> b = \frac{7}{4} </math>
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So: <math> a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow E </math>
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== See Also ==
 
== See Also ==
*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=11|num-a=13}}
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* [[Line | Lines]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:17, 4 July 2013

Problem

The lines $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$ intersect at the point $(1,2)$. What is $a+b$?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4}$

Solution

Since $(1,2)$ is a solution to both equations, plugging in $x=1$ and $y=2$ will give the values of $a$ and $b$.

$1 = \frac{1}{4} \cdot 2 + a$

$a = \frac{1}{2}$

$2 = \frac{1}{4} \cdot 1 + b$

$b = \frac{7}{4}$

So: $a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow E$


See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
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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