1950 AMC 12 第 14 题

Question

对于方程组

\begin{aligned} 2x-3y&=8,\\ 6y-4x&=9\text{，} \end{aligned}

For the simultaneous equations

\begin{aligned} 2x-3y&=8,\\ 6y-4x&=9, \end{aligned}

x=4,

y=0

x=0,

y=\dfrac32

x=0,

y=0

无解

There is no solution

有无穷多个解

There are an infinite number of solutions

比较第二个方程的左边与第一个方程的左边

Compare the left side of the second equation with the left side of the first

若将第一个方程乘以

-2

，则第二个方程的右边应为

-16

Multiplying the first equation by

-2

would require the second right side to be

-16

第二个方程的左边是

-2(2x-3y)

。若第一个方程成立，这个式子必须等于

-2\cdot8=-16

，但第二个方程却规定它等于

9

。这个矛盾说明不存在同时满足两个方程的有序数对。

因此，正确答案是 D。

The left side of the second equation is

-2(2x-3y).

If the first equation holds, that expression must equal

-2\cdot8=-16,

but the second equation says it equals

9.

This contradiction means that no ordered pair satisfies both equations.

Thus, the correct answer is D.

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