2006 AMC 12B Problems/Problem 8: Difference between revisions

Revision as of 08:38, 14 November 2007

The lines $x = \frac 14y + a$ and $y = \frac 14x + b$ intersect at the point $(1,2)$ . What is $a + b$ ?

$\text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94$

$4x-4a=y$

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

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

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

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

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

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 == Problem ==
+The lines <math>x = \frac 14y + a</math> and <math>y = \frac 14x + b</math> intersect at the point <math>(1,2)</math>.  What is <math>a + b</math>?
+<math>
+\text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94
+</math>
 == Solution ==
+<math>4x-4a=y</math>
+<math>4x-4a=\frac{1}{4}x+b</math>
+<math>4*1-4a=\frac{1}{4}*1+b=2</math>
+<math>a=\frac{1}{2}</math>
+<math>b=\frac{7}{4}</math>
+<math>a+b=\frac{9}{4} \Rightarrow \text{(E)}</math>
 == See also ==
 * [[2006 AMC 12B Problems]]