Question 1033677
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What are the solutions to the following system of equations? 
 y + x^2 = 3 
 x^2 + 4y^2 = 36
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x^2 = 3-y
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Substitute and solve for "y"::
3-y + 4y^2 = 36
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4y^2 -y - 33 = 0
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(y-3)(4y+11) = 0
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y = 3 or y = -11/4
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If y = 3, x = Sqrt(0) = 0
If y = -11/4, x = sqrt[3+11/4] = sqrt[23/4]
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Cheers,
Stan H.
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