Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the system of nonlinear equations. $$ \begin{array}{r} 2 x^{3}-x^{2}=y \\ y=\frac{1}{2}-x \end{array} $$

Step-by-Step Solution

Answer

The solutions are (0.5, 0), (0, 0.5), and (-0.5, 1).

Step 1: Substitute Equation 2 into Equation 1

From the second equation, we have: \(y = \frac{1}{2} - x\). Substitute this expression for \(y\) into the first equation: \(2x^3 - x^2 = \frac{1}{2} - x\). Now, we have \(2x^3 - x^2 + x = \frac{1}{2}\).

Step 2: Simplify the Equation

Rearrange the terms to form a single equation: \(2x^3 - x^2 + x - \frac{1}{2} = 0\). This is a cubic equation in terms of \(x\).

Step 3: Attempt to Factor the Cubic Equation

Look for rational solutions by testing possible values. Start with \(x=0.5\), since rational zeros often are factors of the constant term. Substituting \(x = 0.5\) gives \(2(0.5)^3 - (0.5)^2 + 0.5 - 0.5 = 0\), confirming \(x = 0.5\) is a solution.

Step 4: Find Corresponding y Value

Plug \(x = 0.5\) back into the second equation: \(y = \frac{1}{2} - 0.5 = 0\). Thus, the solution is \((0.5, 0)\).

Step 5: Check for More Solutions or Confirm Completeness

After dividing the cubic polynomial \(2x^3 - x^2 + x - \frac{1}{2}\) by \(x - 0.5\) using synthetic or polynomial division, check roots of the resulting quadratic. If necessary, use the quadratic formula.

Step 6: Confirm Additional Solutions

The polynomial division yields \(2x^2 + x\), set to zero gives additional solutions \(x = 0\) and \(x = -\frac{1}{2}\). Substituting these back into the equation for \(y\), \(x = 0\) gives \(y = 0.5\) and \(x = -0.5\) gives \(y = 1\). Thus, the full solution set is \((0.5, 0), (0, 0.5), (-0.5, 1)\).

Key Concepts

System of Equations

A system of equations involves solving for multiple variables simultaneously, utilizing multiple equations. In this exercise, the system comprises two equations involving the variables \(x\) and \(y\):

• \(2x^3 - x^2 = y\)
• \(y = \frac{1}{2} - x\)

The goal is to find pairs \((x, y)\) that satisfy both equations.
One effective method to tackle this problem is by using the substitution technique, which allows for simplifying problems with multiple variables by expressing one variable in terms of another.
This can reduce the problem to a single-variable equation.

Cubic Equation

A cubic equation is a polynomial equation of degree three, characterized by the general form \(ax^3 + bx^2 + cx + d = 0\). In our case, after substitution and simplification, the equation becomes: \[2x^3 - x^2 + x - \frac{1}{2} = 0\]Cubic equations may have three real roots, or just one real root and two complex roots.
In tackling cubic equations, methods such as factoring, synthetic division, or the use of the rational root theorem often assist in finding the solutions.
Testing possible rational roots can quickly lead to a solution, as shown in this exercise.

Substitution Method

The substitution method is where you solve one of the equations for one variable and then substitute that expression into the other equation.
This method simplifies the system of equations by transforming it into a single equation with one variable.Here, we substituted the second equation, \(y = \frac{1}{2} - x\), into the first equation: \[2x^3 - x^2 = \frac{1}{2} - x\]After simplifying, we arrive at a single-variable cubic equation:
\[2x^3 - x^2 + x - \frac{1}{2} = 0\]This reduction makes it easier to manage and ultimately solve for \(x\).
Once \(x\) is found, substitute it back into the expression for \(y\) to find the corresponding \(y\) values.

Factoring Polynomials

Factoring polynomials is an essential technique for solving polynomial equations.
It involves expressing the polynomial as a product of simpler polynomials. In the context of the cubic equation:\[2x^3 - x^2 + x - \frac{1}{2} = 0\]The first step involved testing potential rational roots, leading to the discovery that \(x = 0.5\) is a valid root.
Once a root is identified, synthetic or long division helps to factor out that root from the polynomial.
This process results in a quotient that is typically a simpler polynomial, which can then be solved further:For the remaining equation \(2x^2 + x = 0\), factor further by setting:\[x(2x+1) = 0\]This results in solutions \(x = 0\) and \(x = -\frac{1}{2}\).
Each solution is then substituted back to find corresponding \(y\) values, ensuring the comprehensive set of solutions is found for the original system.