AMC12 2006 B
AMC12 2006 B · Q8
AMC12 2006 B · Q8. It mainly tests Linear equations.
The lines $x = \frac 14y + a$ and $y = \frac 14x + b$ intersect at the point $(1,2)$. What is $a + b$?
直线 $x = \frac 14y + a$ 和 $y = \frac 14x + b$ 相交于点 $(1,2)$。$a + b$ 等于多少？
(A)
0
0
(B)
\frac{3}{4}
\frac{3}{4}
(C)
1
1
(D)
2
2
(E)
\frac{9}{4}
\frac{9}{4}
Answer
Correct choice: (E)
正确答案：（E）
Solution
$4x-4a=y$
$4x-4a=\frac{1}{4}x+b$
$4\cdot1-4a=\frac{1}{4}\cdot1+b=2$
$a=\frac{1}{2}$
$b=\frac{7}{4}$
$a+b=\frac{9}{4} \Rightarrow \fbox{(E)}$
$4x-4a=y$
$4x-4a=\frac{1}{4}x+b$
$4\cdot1-4a=\frac{1}{4}\cdot1+b=2$
$a=\frac{1}{2}$
$b=\frac{7}{4}$
$a+b=\frac{9}{4} \Rightarrow \fbox{(E)}$
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