Precalculus

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Please help me.
Solve the system algebraically.
16)y=x^3+x^2
y=3x^2

  • Precalculus -

    since 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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