Difference between revisions of "2006 AMC 10B Problems/Problem 12"

Revision as of 14:02, 26 January 2022

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

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

@@ Line 2: / Line 2: @@
 The lines <math> x=\frac{1}{4}y+a </math> and <math> y=\frac{1}{4}x+b </math> intersect at the point <math> (1,2) </math>. What is <math> a+b </math>?
-<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4} </math>
+<math> \textbf{(A) } 0\qquad \textbf{(B) } \frac{3}{4}\qquad \textbf{(C) } 1\qquad \textbf{(D) } 2\qquad \textbf{(E) } \frac{9}{4} </math>
 == Solution ==