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\par{John Doe \newline{} Math 587 Assignment 2 \newline{} March 3, 2009
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\item  At what point does the line \(2 x - y \, = \, 3\) intersect the line \(x + 2y \, = \, -1\)? \  \par{\emph{Solution.} The point of intersection may be obtained by solving the two equations simultaneously.    For this multiply the first equation by \(2\) and add that to the second equation obtaining the equation \[ 5 x \ = \  5 \ . \] Thus, \(x \, = \, 1\), and, using either of the two original equations, one finds \(y \, = \, -1\).    The required point is \((1, -1)\).   
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\item  Find all solutions of the quadratic equation \(x^{2} - x - 12 \, = \, 0\).   \par{\emph{Solution.} The well known formula for solution of the quadratic equation \(ax^{2} + b x + c \, = \, 0\) is \[ x \ = \  \frac{-b \pm{} \sqrt{b^{2} - 4ac}}{2a} \ . \] In this case one finds \[ x \ = \  \frac{1 \pm{} \sqrt{1 -4(1)(-12)}}{2}      \ = \  \frac{1 \pm{} 7}{2}      \ = \  4 \ \mbox{or} \ -3 \ .\] 
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