Problem 72 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

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 1: 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.

Step 2: Collect like terms

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

Step 3: Solve for x

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

Step 4: 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 \(x = -1\), \(y = -1\).