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If: x^2+y^2+4x+6y -40 = 0 and x -y = 10

Then by rearranging: x = 10+y and 2y^2+30y+100 = 0

Solving the above quadratic equation: y = -10 and y = -5

Points of intersection by substitution are: (0, -10) and (5, -5)

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