1950 AMC 12 第 6 题

Question

满足下列方程组的

y

值

\begin{aligned} 2x^2+6x+5y+1&=0,\\ 2x+y+3&=0 \end{aligned}

可以通过求解下列方程得到：

The values of

y

which will satisfy the equations

\begin{aligned} 2x^2+6x+5y+1&=0,\\ 2x+y+3&=0 \end{aligned}

may be found by solving:

y^2+14y-7=0

y^2+8y+1=0

y^2+10y-7=0

y^2+y-12=0

以上方程均不正确

None of the above equations

将线性方程中的

x

用

y

表示

Solve the linear equation for

x

in terms of

y

将

x=-\frac{(y+3)}{2}

代入第一个方程，再消去分母

Substitute

x=-\frac{(y+3)}{2}

into the first equation and clear the denominator

由第二个方程可得

x=-\tfrac{y+3}{2}

。将其代入第一个方程并乘以

2

，得到

\begin{aligned} &(y+3)^2-6(y+3)\\ &\quad+10y+2=0,\\ &y^2+10y-7=0 \end{aligned}\text{。}

因此，正确答案是 C。

From the second equation,

x=-\tfrac{y+3}{2}.

Substituting into the first equation and multiplying by

2

gives

\begin{aligned} &(y+3)^2-6(y+3)\\ &\quad+10y+2=0,\\ &y^2+10y-7=0. \end{aligned}

Thus, the correct answer is C.

其他年份的第 6 题