AMC10 2006 B

AMC10 2006 B · Q12

AMC10 2006 B · Q12. It mainly tests Linear equations, Systems of equations.

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

直线 $x = \frac{1}{4}y + a$ 和 $y = \frac{1}{4}x + b$ 相交于点 $(1,2)$。求 $a+b$。

(A) 0 0

(B) $\frac{3}{4}$ $\frac{3}{4}$

(D) 2 2

(E) $\frac{9}{4}$ $\frac{9}{4}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Substituting $x = 1$ and $y = 2$ into the equations gives $$1 = \frac{2}{4} + a \quad \text{and} \quad 2 = \frac{1}{4} + b.$$ It follows that $$a + b = \left(1 - \frac{2}{4}\right) + \left(2 - \frac{1}{4}\right) = 3 - \frac{3}{4} = \frac{9}{4}.$$

将 $x = 1$ 和 $y = 2$ 代入方程组得 $$1 = \frac{2}{4} + a \quad \text{和} \quad 2 = \frac{1}{4} + b。$$ 因此 $$a + b = \left(1 - \frac{2}{4}\right) + \left(2 - \frac{1}{4}\right) = 3 - \frac{3}{4} = \frac{9}{4}。$$

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