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 four and simplifying, we find:

$2*(2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1)=2*0$ $(y+3)^2 -6y-18+10y+2=0$ $y^2+6y+9-6y-18+10y+2=0$ $y^2+10y-7=0$

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

1950 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1950 AHSME Problems/Problem 6

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