The values of y which will satisfy the equations 2x^2+6x+5y+1&=0, 2x+y+3&=0 may be found by solving:

(E) None of the above equations

1950 AMC 12 Problem 6

Attempt Problem 6 of the 1950 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1950 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

None of the above equations

Solve the linear equation for

x

in terms of

y

Substitute

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

into the first equation and clear the denominator

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.