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December 19, 2014

December 19, 2014

Posted by **Phoebe** on Tuesday, March 5, 2013 at 6:33pm.

Solve the system algebraically.

16)y=x^3+x^2

y=3x^2

- Precalculus -
**Dr. Jane**, Tuesday, March 5, 2013 at 7:41pmsince both equations = y, you can set each equation next to each other.

x^3 +x^2 = 3x^2

Subtract 3x^2 from each side.

x^3 -2x^2 = 0

Factor out the GCF

2x^2 (x - 1) = 0

2x^2 = 0 x - 1 = 0

solve for x.

then find y in one of the original equations once you know x.

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