Problem 71 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The solution set for the system of equations is: (0,1) and (1,0)

Step 1: Set the equations equal to each other

Since both equations equal y, set them equal to each other. So now, the equation to solve would be \(x^{3}-2x^{2}+1 = 1-x^{2}\)

Step 2: Simplify Equation

Combine like terms and simplify the equation acquired in step 1. This leads to \(x^{3}-2x^{2}+x^{2}-1+1 = 0\) which further simplifies to \(x^{3}-x^{2} = 0\)

Step 3: Factor out x in the equation

The equation \(x^{3}-x^{2}=0\) can be factored into \(x^{2}(x-1) = 0\)

Step 4: Solve for x

Set each factor equal to zero and solve for x values. The equation \(x^{2} = 0\) yields an x-value of 0 and the equation \(x-1 = 0\) yields an x-value of 1. This means the solution set for the variable x is \{0,1\}

Step 5: Solve for y

Substitute x=0 and x=1 into one of the original equations to find the corresponding y values. Using the equation \(y=1-x^{2}\), when x=0, y=1, and when x=1, y=0. Therefore, the solution set for the system of equations is: (0,1) and (1,0)

Key Concepts

Graphical Solution

To solve a system of equations graphically means to plot both equations on a graph and find their points of intersection. These intersections are the solutions to the system, as they represent the points where both equations have the same values for x and y.
In our exercise, the equations are:

• \( y = x^3 - 2x^2 + 1 \)
• \( y = 1 - x^2 \)

Plotting these on a coordinate grid helps visualize their relationship. Where they intersect, the values of x and y satisfy both equations simultaneously. For this specific problem, the graphical method shows intersections at the points (0,1) and (1,0). These points form the solution set.
This method is very intuitive because you can see the exact nature of each function's curve and how they relate to one another.

Algebraic Solution

An algebraic solution involves manipulating the equations using algebra to find the precise values of x and y that satisfy both equations. This approach is particularly useful when the graphical representation may be difficult to interpret accurately, especially if intersections occur at complex or irrational numbers.
In the original problem:

• We start by equating the two equations because they both equal \(y\): \( x^3 - 2x^2 + 1 = 1 - x^2 \)
• Simplify the resulting equation to make it easier to manipulate further: \( x^3 - x^2 = 0 \)

This form is much simpler and makes it possible to apply other algebraic methods, such as factoring, to find solutions.

Factoring

Factoring is a powerful algebraic tool used to simplify polynomial equations and find their roots. In our context, factoring breaks down the equation into simpler parts that can be individually solved.
From the simplified equation \( x^3 - x^2 = 0 \), factoring out \( x^2 \) gives us: \( x^2(x - 1) = 0 \).
Each factor is solved separately:

• \( x^2 = 0 \) leads to \( x = 0 \)
• \( x - 1 = 0 \) leads to \( x = 1 \)

By solving each factor, you determine the x-values that make the entire equation true. These values are key components of the solution set.

Solution Set

The solution set of a system of equations is the collection of all possible values for x and y that simultaneously satisfy all equations in the system.
For the given system, we found the x-values using algebraic methods and factoring: \{0, 1\}.
Then, we substitute these x-values back into one of the original equations to find the corresponding y-values:

• For \( x = 0 \), substituting into \( y = 1 - x^2 \) gives \( y = 1 \)
• For \( x = 1 \), substituting into \( y = 1 - x^2 \) gives \( y = 0 \)

Therefore, the solution set that satisfies both original equations is \((0, 1) \) and \((1, 0)\).
This set is where the graphical and algebraic methods converge, ensuring thorough comprehension of the solution by different methods.