William W. answered • 12/02/20

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x^{2} + 2y = 12

x^{2} = 12 - 2y

x^{2} + y^{2} = 36

x^{2} = 36 - y^{2}

So, since x^{2} = 12 - 2y and x^{2} = 36 - y^{2} then 12 - 2y must equal 36 - y^{2}

12 - 2y = 36 - y^{2}

y^{2} - 2y - 24 = 0

(y - 6)(y + 4) = 0

y = 6 and y = -4

Since x^{2} + 2y = 12 then, first plugging in y = 6, we get:

x^{2} + 2(6) = 12

x^{2} = 0

So, (0, 6) is an intersection point.

The plugging in y = -4:

x^{2} + 2(-4) = 12

x^{2} = 20

x = ± √20 = ±2√5

So (2√5, -4) is an intersection point and (-2√5, -4) is an intersection point.

Another method that often work nicely is Elimination:

Multiply both sides of x^{2} + 2y = 12 by negative 1 to get: -x^{2} - 2y = -12 then place it over the other equation and add the two equations together to eliminate "x":

-x^{2} - 2y = -12

x^{2} + y^{2} = 36

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y^{2} - 2y = 24

The solve as noted above.