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AHSME 1950- the original problems

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July 17, 2008

Problem Statement

The values of y which will satisfy the equations 2x^2\plus{}6x\plus{}5y\plus{}1\equal{}0, 2x\plus{}y\plus{}3\equal{}0 may be found by solving:
<spanclass=′latex−bold′>(A)</span> y2+14y−7=0<spanclass=′latex−bold′>(B)</span> y2+8y+1=0<spanclass=′latex−bold′>(C)</span> y2+10y−7=0<spanclass=′latex−bold′>(D)</span> y2+y−12=0<spanclass=′latex−bold′>(E)</span> None of these equations<span class='latex-bold'>(A)</span>\ y^2+14y-7=0 \qquad <span class='latex-bold'>(B)</span>\ y^2+8y+1=0 \qquad <span class='latex-bold'>(C)</span>\ y^2+10y-7=0 \qquad <span class='latex-bold'>(D)</span>\ y^2+y-12=0 \qquad <span class='latex-bold'>(E)</span>\ \text{None of these equations}
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