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X-y=4+y Xy=12 Find all solutions of the system of equations
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\[x-y=4+y\iff x=4+2y\]
substitute in to the second equation, get\[(4+2y)y=12\]
multiply out, get \[2y^2+4y=12\] or \[y^2+2y=6\] or \[y^2+2y-6=0\]
this does not factor, so you will need to use either the quadratic formula, or complete the square, to find \(y\) since \(2\) is even, completing the square is easiest \[y^2+2y=6\] \[(y+1)^2=7\] \[y+1=\pm\sqrt7\] \[y=-1\pm\sqrt7\]
i assume the original problem is \[x-y=4+y\\xy=12\] as posted, right?
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