1950 AHSME Problems/Problem 6

Problem

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

$\textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations}$

Solution

If we solve the second equation for $x$ in terms of $y$, we find $x=-\dfrac{y+3}{2}$ which we can substitute to find:

\[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0\]

Multiplying by two and simplifying, we find:

\begin{align*} 2\cdot[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1]&=2\cdot 0\\ (y+3)^2 -6y-18+10y+2&=0\\ y^2+6y+9-6y-18+10y+2&=0\\ y^2+10y-7&=0 \end{align*}

Therefore the answer is $\boxed{\textbf{(C)}\ y^{2}+10y-7=0}$

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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