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

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== Problem == | == Problem == | ||

+ | == Problem 12 == | ||

+ | 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>? | ||

+ | |||

+ | <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>. | ||

+ | |||

+ | <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> | ||

+ | |||

== See Also == | == See Also == | ||

*[[2006 AMC 10B Problems]] | *[[2006 AMC 10B Problems]] |

## Revision as of 20:24, 13 July 2006

## Contents

## Problem

## Problem 12

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: