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

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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>?
 
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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<math> \textbf{(A) } 0\qquad \textbf{(B) } \frac{3}{4}\qquad \textbf{(C) } 1\qquad \textbf{(D) } 2\qquad \textbf{(E) } \frac{9}{4} </math>
  
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== Solution ==
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== Solution 1 ==
 
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>.  
 
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> b = \frac{7}{4} </math>
 
<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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So <math> a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}</math>.
  
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== Solution 2 ==
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Substituting <math>x=1</math> and <math>y=2</math> into the equations gives <math>1=\frac{2}{4}+a\quad\text{and}\quad 2=\frac{1}{4}+b.
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</math> It follows that <math>a+b=\left(1-\frac{2}{4}\right)+\left(2-\frac{1}{4}\right)=3 - \frac{3}{4}=\boxed{\frac{9}{4}}.</math>
  
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== See Also ==
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== Solution 3==
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*[[2006 AMC 10B Problems]]
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Because <math>a=x- \frac{y}{4}\quad\text{and}\quad b=y-\frac{x}{4}
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\quad\text{we have}\quad a +b = \frac{3}{4}(x+y).</math> Since <math>x=1</math> when <math>y=2</math>, this implies that <math>a+b = \frac{3}{4}(1 + 2) = \boxed{\frac{9}{4}}</math>.
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== Video Solution ==
  
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*[[2006 AMC 10B Problems/Problem 11|Previous Problem]]
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https://www.youtube.com/watch?v=q1mxhiyciSk  ~David
  
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*[[2006 AMC 10B Problems/Problem 13|Next Problem]]
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== See Also ==
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{{AMC10 box|year=2006|ab=B|num-b=11|num-a=13}}
  
 
* [[Line | Lines]]
 
* [[Line | Lines]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:07, 27 November 2023

Problem

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

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

Solution 1

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 $a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}$.

Solution 2

Substituting $x=1$ and $y=2$ into the equations gives $1=\frac{2}{4}+a\quad\text{and}\quad 2=\frac{1}{4}+b.$ It follows that $a+b=\left(1-\frac{2}{4}\right)+\left(2-\frac{1}{4}\right)=3 - \frac{3}{4}=\boxed{\frac{9}{4}}.$

Solution 3

Because $a=x- \frac{y}{4}\quad\text{and}\quad b=y-\frac{x}{4} \quad\text{we have}\quad a +b = \frac{3}{4}(x+y).$ Since $x=1$ when $y=2$, this implies that $a+b = \frac{3}{4}(1 + 2) = \boxed{\frac{9}{4}}$.

Video Solution

https://www.youtube.com/watch?v=q1mxhiyciSk ~David

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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