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If: 3x-y = 5 and 2x2+y2 = 129

Then: 3x-y = 5 => y = 3x-5

And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0

Using the quadratic equation formula: x = 52/11 and x = -2

By substitution points of intersection are: (52/11, 101/11) and (-2, -11)

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Q: What are the points of intersection of the line 3x -y equals 5 with 2x squared plus y squared equals 129 showing work?
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