Difference between revisions of "2006 AMC 12B Problems/Problem 8"

Revision as of 09:35, 16 September 2017

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

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

$4x-4a=y$

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

$4\cdot1-4a=\frac{1}{4}\cdot1+b=2$

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

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

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

Add both equations:

$[x=\frac{1}{4}y+a]+[y=\frac{1}{4}x+b]$

Simplify:

$\frac{4}{4}(x+y)=\frac{1}{4}(x+y)+(a+b)$

Isolate our solution:

$\frac{3}{4}(x+y)=a+b$

Substitute the point of intersection $[x=1, y=2]$

$a+b=\frac{3}{4}\cdot3=\frac{9}{4} \Rightarrow \text{(E)}$

Plugging in $(1,2)$ into the first equation, and solving for $a$ we get $a$ as $\frac{1}{2}$ .

Doing the same for the second equation for the second equation, we get $b$ as $\frac{7}{4}$

Adding $a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow \text{(E)}$

2006 AMC 12B (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

@@ Line 35: / Line 35: @@
 <math>a+b=\frac{3}{4}\cdot3=\frac{9}{4} \Rightarrow \text{(E)}</math>
+==Solution 3==
+Plugging in <math>(1,2)</math> into the first equation, and solving for <math>a</math> we get <math>a</math> as <math>\frac{1}{2}</math>.
+Doing the same for the second equation for the second equation, we get <math>b</math> as <math>\frac{7}{4}</math>
+Adding <math>a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow \text{(E)}</math>
 == See also ==
 {{AMC12 box|year=2006|ab=B|num-b=7|num-a=9}}
 {{MAA Notice}}