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If: x = 2y-2 and x^2 = y^2+7

Then: (2y-2)^2 = y^2+7 => 4y^2-8y+4 = y^2+7 => 3y^2-8y-3= 0

Solving the above quadratic equation: y = -1/3 or y = 3

Solutions by substitution: when y=-1/3 then x=-8/3 and when y=3 then x=4

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