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2006 AMC 12B Problems/Problem 8

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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

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

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

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

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

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

See also

2006 AMC 12B (Problems • Resources)
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12B_Problems/Problem_8"

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