# 2006 AMC 12B Problems/Problem 8

## Problem

The lines $x = \frac 14y + a$ and $y = \frac 14x + b$ intersect at the point $(1,2)$. What is $a + b$? $\text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94$

## Solution 1 $4x-4a=y$ $4x-4a=\frac{1}{4}x+b$ $4\cdot1-4a=\frac{1}{4}\cdot1+b=2$ $a=\frac{1}{2}$ $b=\frac{7}{4}$ $a+b=\frac{9}{4} \Rightarrow \text{(E)}$

## Solution 2 $\begin{cases}x=\frac{1}{4}y+a\\ y=\frac{1}{4}x+b \end{cases}$

Simplify: $\frac{4}{4}(x+y)=\frac{1}{4}(x+y)+(a+b)$

Isolate our solution: $\frac{3}{4}(x+y)=a+b$

Substitute the point of intersection $[x=1, y=2]$ $a+b=\frac{3}{4}\cdot3=\frac{9}{4} \Rightarrow \text{(E)}$

## Solution 3

Plugging in $(1,2)$ into the first equation, and solving for $a$ we get $a$ as $\frac{1}{2}$.

Doing the same for the second equation for the second equation, we get $b$ as $\frac{7}{4}$

Adding $a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow \text{(E)}$

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