Problem 20 - College Algebra

Question

Use any method to solve the system of nonlinear equations. $$ \begin{aligned} 2 x^{3}-x^{2} &=y \\ y &=\frac{1}{2}-x \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( (x, y) = \left(\frac{1}{2}, 0\right) \).

Step 1: Understand the System of Equations

The system consists of two equations: 1. The first equation is \( 2x^3 - x^2 = y \).2. The second equation is \( y = \frac{1}{2} - x \).We need to find values of \( x \) and \( y \) that satisfy both equations simultaneously.

Step 2: Substitute y in the First Equation

Substitute \( y = \frac{1}{2} - x \) into the first equation. This gives us the equation:\[ 2x^3 - x^2 = \frac{1}{2} - x \].

Step 3: Rearrange the Equation

Bring all terms to one side of the equation to set the equation to zero:\[ 2x^3 - x^2 + x - \frac{1}{2} = 0 \].

Step 4: Solve the Polynomial Equation

To solve \( 2x^3 - x^2 + x - \frac{1}{2} = 0 \), you can try methods like factoring or numerical methods. Here, let us check a potential root by substitution. Test some rational values, such as \( x = \frac{1}{2} \):\[ 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + \frac{1}{2} - \frac{1}{2} = 0 \], which simplifies to \[ \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{1}{2} = 0 \]. Hence, \( x = \frac{1}{2} \) is a solution.

Step 5: Find Corresponding y Value

Using \( x = \frac{1}{2} \), substitute back into \( y = \frac{1}{2} - x \):\[ y = \frac{1}{2} - \frac{1}{2} = 0 \]. Thus, \((x, y) = \left(\frac{1}{2}, 0\right)\) is a solution.

Key Concepts

Nonlinear Equations

Nonlinear equations are mathematical expressions in which the variables are raised to a power other than one, or are multiplied together, which results in graphs that aren't straight lines. These equations can take many forms, such as quadratics, cubics, or higher-degree polynomials. In our exercise, the first equation, \( 2x^3 - x^2 = y \), is nonlinear because it involves terms with variables raised to the power of two and three. Nonlinear equations can intersect at multiple points or potentially have no points in common, which adds complexity to their analysis. Understanding the behavior of these equations often requires methods such as substitution or graphing to find solutions.

Solving Systems of Equations

Solving systems of equations involves finding values of variables that satisfy all equations in the system simultaneously. For our given system, we have two equations: one nonlinear and one linear.

• The first equation is nonlinear: \( 2x^3 - x^2 = y \).
• The second is linear: \( y = \frac{1}{2} - x \).

The goal is to identify values of \( x \) and \( y \) that satisfy both equations at the same time. Steps typically involve rearranging and substituting one equation into the other to eliminate a variable, which simplifies the process to finding solutions to a single equation as done in the original problem.

Polynomial Equations

Polynomial equations are expressions that consist of variables and coefficients involving operations of addition, subtraction, and multiplication, along with non-negative integer exponents of variables. Our main polynomial equation from the exercise, \( 2x^3 - x^2 + x - \frac{1}{2} = 0 \), is a cubic polynomial due to the highest power being three. Solving these equations can involve factoring, using the quadratic formula for simpler polynomials, or applying numerical techniques like the Rational Root Theorem. Determining solutions often includes checking possible roots, similar to what we observed with testing \( x = \frac{1}{2} \) in the exercise, to verify if they satisfy the polynomial equation.

Algebraic Substitution

Algebraic substitution is a technique used to simplify systems of equations, particularly useful when one equation is solved for one variable. It involves replacing a variable in one equation with an equivalent expression from another equation. In the step-by-step solution, substitution was used by replacing \( y \) in the first equation with \( \frac{1}{2} - x \) from the second equation.

• This reduces the system to a single equation: \( 2x^3 - x^2 + x - \frac{1}{2} = 0 \).
• By solving this polynomial equation, we found the \( x \) value that satisfies both original equations.

This method is powerful as it breaks a complex system into a more manageable form, ultimately allowing for easier identification of solutions.