Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the system of nonlinear equations. $$ \begin{array}{c} 2 x^{3}-x^{2}=y \\ x^{2}+y=0 \end{array} $$

Step-by-Step Solution

Answer

The solution is \((0, 0)\).

Step 1: Express y from the second equation

Start by solving the second equation for \( y \). The equation is \( x^2 + y = 0 \). By isolating \( y \), we get \( y = -x^2 \).

Step 2: Substitute y into the first equation

Use the expression for \( y \) from Step 1 and substitute it into the first equation. The first equation is \( 2x^3 - x^2 = y \), and substituting \( y = -x^2 \) gives us \( 2x^3 - x^2 = -x^2 \).

Step 3: Simplify the equation

Simplify the equation \( 2x^3 - x^2 = -x^2 \) by adding \( x^2 \) to both sides, resulting in \( 2x^3 = 0 \).

Step 4: Solve for x

Solve the equation \( 2x^3 = 0 \) for \( x \). Dividing both sides by 2 gives \( x^3 = 0 \). Taking the cube root of both sides, we find \( x = 0 \).

Step 5: Find y using x-value

Substitute \( x = 0 \) back into the expression \( y = -x^2 \) from Step 1. This gives \( y = -0^2 = 0 \).

Step 6: Conclude the solution

The solution to the system of equations is the point \((x, y) = (0, 0)\).

Key Concepts

Solving Systems of Equations

In mathematics, a system of equations is a collection of two or more equations with the same set of unknowns. Solving these systems can sometimes be straightforward, but when the equations are nonlinear, things might get tricky. In our example, we have a system of two equations:

• \(2x^3 - x^2 = y\)
• \(x^2 + y = 0\)

The approach for solving systems of equations depends on the nature and complexity of the equations. You may choose various methods like graphing, substitution, or elimination.
In this case, nonlinear equations demand careful manipulation to find their common solution. The aim is to find values for \(x\) and \(y\) that satisfy both equations simultaneously.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions and equations using algebraic rules and operations, such as addition, subtraction, multiplication, and division. It is often employed to isolate variables or express a variable in terms of another.
In the given system, algebraic manipulation helps to first solve one of the equations to express \(y\) in terms of \(x\). Taking the second equation, \(x^2 + y = 0\), and solving it gives:

• \(y = -x^2\)

This expression for \(y\) can then be substituted into the first equation, aiding in finding a common solution.
Without algebraic manipulation, it'd be quite challenging to solve nonlinear systems of equations.

Substitution Method

The substitution method is a powerful way of solving systems of equations, especially when one of the equations can be easily solved for one of the variables.
For our system of equations, solving the second equation for \(y\) gives \(y = -x^2\). This expression is substituted back into the first equation, transforming it into a single variable equation:

• \(2x^3 - x^2 = -x^2\)

The substitution simplifies the problem by reducing two equations into one, making it easier to solve for \(x\). Once \(x\) is determined, the known relation \(y = -x^2\) helps find \(y\).
Substitution is particularly effective when dealing with nonlinear equations since it avoids complex elimination processes.

Solution of Equations

Solving equations is about finding the numerical values of variables that make the equation true. In the case of nonlinear equations, finding the solution often requires careful step-by-step approaches.
For our system, after substituting and simplifying, we are left with the equation:

• \(2x^3 = 0\)

Solving for \(x\) involves dividing by the coefficient of \(x^3\) and then taking roots where necessary. Here, dividing by 2 and taking the cube root gives:

• \(x = 0\)

Substitute \(x = 0\) back into the expression \(y = -x^2\), resulting in \(y = 0\). This gives the solution \((x, y) = (0, 0)\), a point where both equations are satisfied.
It's important to meticulously verify that the found solution satisfies both of the original equations. This ensures the validity of the obtained solution.