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Difference between revisions of "1950 AHSME Problems/Problem 6"

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Revision as of 13:28, 17 April 2012

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 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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