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Chapter 8: Problem 72

Solve the system graphically or algebraically. Explain your choice of method. $$\left\\{\begin{array}{l} y=x^{3}-2 x^{2}+x-1 \\ y=-x^{2}+3 x-1 \end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are (0, -1), (2, 1) and (-1, -1). This means these are the points where the graphs of the two equations intersect.

Step by step solution

01

Set the two equations equal to each other

We have two equations as \(y\), we could set them equal to each other, which is: \(x^{3} - 2x^{2} + x - 1 = -x^{2} + 3x - 1\). Simplify the equation by adding \(x^{2}\) and subtracting \(3x\) to both sides, we take \(-1\) to the other side.
02

Collect like terms

Combine like terms to simplify the equation. Then get: \(x^{3} - x^{2} - 2x = 0\). Factor from the equation: \(x(x - 2)(x + 1) = 0\)
03

Solve for x

Set each factor equal to zero and solve for x. This gives us \(x = 0\), \(x = 2\), and \(x = -1\).
04

Find y-coordinate

Substitute each solution for x into one of the original equations (either would yield the same result) to solve for the corresponding y-coordinate: when \(x = 0\), \(y = -1\); when \(x = 2\), \(y = 1\); and when \(x = -1\), \(y = -1\).

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