2006 AMC 10B Problems/Problem 12

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

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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