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

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.