Question

设 $x$ 和 $y$ 是非零实数，且满足 $\frac{3x+y}{x-3y}=-2\text{。}$ 求下式的值： $\frac{x+3y}{3x-y}\text{。}$

Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2.$ What is the value of $\frac{x+3y}{3x-y}?$

$-3$

$-1$

$1$

$2$

$3$

将给定方程清除分母。

Clear the denominator in the given equation

这个方程很快会推出 $x$ 与 $y$ 的关系。

The equation quickly forces a relation between $x$ and $y$

由 $\frac{3x+y}{x-3y}=-2\text{，}$ 两边同乘分母，得到 $3x+y = 6y-2x\text{。}$ 整理可得 $x=y$ 。

因此 $\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2\text{。}$

所以正确答案是 D。

Given that $\frac{3x+y}{x-3y}=-2,$ we can multiply by the denominator to get $3x+y = 6y-2x.$ Solving, we can see that $x=y.$

Therefore, $\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2.$

Thus, the correct answer is D .