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2006 AMC 10B Problems/Problem 12

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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?

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

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

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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