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

Then: (2y-2)^2 = y^2+7

So: (2y-2)(2y-2) = y^2+7

It follows: 4y^2-y^2-8y+4-7 = 0 => 3y^2-8y-3 = 0

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

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

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