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

$4x-4a=y$

$4x-4a=\frac{1}{4}x+b$

$4*1-4a=\frac{1}{4}*1+b=2$

$a=\frac{1}{2}$

$b=\frac{7}{4}$

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

See also

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

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