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

Add both equations:

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

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

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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