2006 AMC 10B Problems/Problem 12: Difference between revisions

Revision as of 23:55, 24 November 2023

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

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

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 $a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}$ .

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 10 Problems and Solutions

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 So <math> a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}</math>.
+== Video Solution ==
+https://www.youtube.com/watch?v=q1mxhiyciSk
 == See Also ==