`y = x^3 - 2x^2 + x - 1, y = -x^2 + 3x - 1` Solve the system graphically or algebraically. Explain your choice of method.

Let's solve this system algebraically. The reasons are: a) not to draw a graph and b) highest accuracy.

We have

`x^3-2x^2+x-1=-x^2+3x-1,` or

`x^3-x^2-2x=0.`

There is one root x=0 and the remaining equation is

`x^2-x-2=0.`

Its roots are x=-1 and x=2.

The corresponding y's are y=-1, y=-5 and y=1.

The answer: 1) x=0, y=-1; 2) x=-1, y=-5; 3) x=2, y=1.