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

Revision as of 21:24, 13 July 2006

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

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

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

@@ Line 1: / Line 1: @@
 == Problem ==
+== Problem 12 ==
+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>
 == Solution ==
+Since <math>(1,2)</math> is a solution to both equations, plugging in <math>x=1</math> and <math>y=2</math> will give the values of <math>a</math> and <math>b</math>.
+<math> 1 = \frac{1}{4} \cdot 2 + a </math>
+<math> a = \frac{1}{2} </math>
+<math> 2 = \frac{1}{4} \cdot 1 + b </math>
+<math> b = \frac{7}{4} </math>
+So: <math> a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow E </math>
 == See Also ==
 *[[2006 AMC 10B Problems]]