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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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8y ago
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8y ago

Since the first equation is already solved for "x", you can replace "x" by "2y-2" in the second equation and continue from there.

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Q: What are the solutions to the simultaneous equations of x equals 2y -2 and x squared equals y squared plus 7 showing work?
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