2006 AMC 10B Problems/Problem 12

Revision as of 20:24, 13 July 2006 by Xantos C. Guin (talk | contribs) (Added problem and solution)

Problem

Problem 12

The lines $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$ intersect at the point $(1,2)$. What is $a+b$?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4}$

Solution

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

See Also

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