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

Then: x = 5/3+y/3 and so 2(5/3+y/3)2+y2 = 129 => 50+20y+2y2+9y2 = 1161

Therefore: 11y2+20y-1111 = 0

Solving the quadratic equation: y = -11 or y = 101/11

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

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