Problem 36 - College Algebra

Question

Solve each system by the method of your choice. $$\left\\{\begin{array}{c} x^{3}+y-0 \\ 2 x^{2}-y-0 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions to the system of equations are (0,0) and (-2,8).

Step 1: Express y from first equation

Rearranging the first equation \(x^{3} + y = 0\) to find \(y\), we get \(y = -x^{3}\).

Step 2: Substitute y in second equation

Now, substitute \(y = -x^{3}\) into the second equation which is \(2x^{2} -y = 0\), hence the equation becomes \(2x^{2} - (-x^{3}) = 0\).

Step 3: Solve for x

Solving for \(x\) implies solving \(2x^{2} + x^{3} = 0\) equation. This can be simplified by factoring \(x^{2}(2 + x) = 0\). From this equation, we get two solutions, \(x = 0\) or \(x = -2\).

Step 4: Solve for y

Substitute \(x\) values into \(y = -x^{3}\) to get corresponding \(y\) values. For \(x = 0\), \(y = 0\) and for \(x = -2\), \(y = -(-8) = 8\).

Key Concepts

Cubic Equations

Cubic equations are a type of polynomial equation characterized by the highest power of the variable being three. They have the general form of \( ax^3 + bx^2 + cx + d = 0 \), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). Working with cubic equations is crucial because they allow for the modeling of more complex relationships than linear or quadratic equations.

• Cubic equations can have one or three real roots, with the potential for complex numbers in solutions.
• In systems of equations, you often work with cubic terms to find interactions between variables, just like in our original exercise.
• They often require special methods such as factoring or substitution to solve when embedded in a system of equations.

Understanding how cubic equations can be reformulated helps to simplify the solution process and discover hidden relationships between variables.

Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of equations. This approach involves solving one equation for one variable and then substituting this expression into another equation. The goal is to reduce the system into a simpler form that can be solved more easily.
In the provided exercise, we start by expressing \(y\) in terms of \(x\) from the first equation \(x^3 + y = 0\). This gives us \(y = -x^3\). By substituting \(y = -x^3\) into the second equation, we eliminate the \(y\) and focus entirely on \(x\).

• Substitution simplifies systems by reducing the number of variables and equations at one time.
• This method is particularly useful when one equation is easier to manipulate or has a straightforward representation, as seen in our example.
• It's essential to carefully handle substitutions to maintain mathematical equality throughout the process.

Factoring

Factoring is a critical technique in algebra that involves breaking down expressions into simpler multiplicative forms. In solving equations, factoring allows us to find solutions by setting factors equal to zero.
In the original step-by-step solution, we combined the terms in equation \(2x^2 + x^3 = 0\) into a factored form: \(x^2(2 + x) = 0\). This expression indicates that either \(x^2 = 0\) or \(2 + x = 0\), leading to the solutions \(x = 0\) or \(x = -2\).

• Factoring helps identify potential solutions that satisfy the equation.
• A factored equation set to zero can be split into multiple simpler equations.
• Each factor may contribute to the overall solution, hence it's important to identify all possible factors and solve them separately.

Understanding factoring provides insight into the solutions and their corresponding values, allowing us to solve more complex problems.

Algebraic Solutions

Algebraic solutions involve using algebraic methods such as factoring, substitution, and simplification to solve equations. These methods convert complex, unsolvable problems into straightforward ones.
Our exercise showcases solving a system of equations by first expressing one variable in terms of another using substitution. Then, we solved the derived cubic equation through factoring.

• Combining multiple algebraic techniques often provides the best path to a solution.
• Algebraic solutions emphasize understanding relationships between expressions and variables.
• They ensure solutions are derived logically, maintaining consistency and accuracy.

When dealing with systems of equations, these strategies simplify complex computations and yield clear, defined results for the variables involved.