Problem 6 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{aligned} x-y^{3} &=1 \\ 2 x &=9 y^{2}+2 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{737}{8}\) and \(y = \frac{9}{2}\).

Step 1: Solve First Equation for x

Start by isolating \(x\) in the first equation, which is \(x - y^3 = 1\). Add \(y^3\) to both sides to get \(x = y^3 + 1\).

Step 2: Substitute x into Second Equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation: \(2(y^3 + 1) = 9y^2 + 2\). Expand to get \(2y^3 + 2 = 9y^2 + 2\).

Step 3: Simplify the Equation

Subtract \(2\) from both sides of the equation obtained in Step 2 to simplify: \(2y^3 = 9y^2\).

Step 4: Factor the Simplified Equation

Factor the equation \(2y^3 = 9y^2\). If possible, divide the entire equation by \(y^2\) (assuming \(y eq 0\)) to get \(2y = 9\).

Step 5: Solve for y

Solve \(2y = 9\) by dividing both sides by 2: \(y = \frac{9}{2}\).

Step 6: Solve for x Using y Value

Substitute \(y = \frac{9}{2}\) back into the expression for \(x\) found in Step 1: \(x = (\frac{9}{2})^3 + 1\). Calculate \((\frac{9}{2})^3\) to find \(x = \frac{729}{8} + 1\). Convert 1 to \(\frac{8}{8}\), and add to get \(x = \frac{737}{8}\).

Step 7: Verify Solution

Substitute the values \(x = \frac{737}{8}\) and \(y = \frac{9}{2}\) back into the original equations to verify they satisfy both equations. This confirms the solution is correct.

Key Concepts

Substitution Method

The substitution method is a popular technique used for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This method works well when one equation is easily solvable for one variable.
To start with substitution, isolate one variable in one of the equations. For example, if we have the equation \(x - y^3 = 1\), we can solve for \(x\), giving us \(x = y^3 + 1\).
Next, take this expression for \(x\) and substitute it into the other equation, \(2x = 9y^2 + 2\). This changes the second equation into one involving only the variable \(y\), making it much simpler to solve. By substituting, we effectively reduce the system of equations to a single polynomial equation in one variable.

Factoring Equations

Factoring equations is an essential step when simplifying expressions or solving polynomial equations, especially after substitution. When we reach an equation such as \(2y^3 = 9y^2\), factoring becomes crucial.
First, we notice common factors that can be divided out. In this case, both sides of the equation have terms involving \(y\). We factor out \(y^2\) from both sides, assuming \(y eq 0\), resulting in \(2y = 9\).
By performing this factoring step, we've simplified a cubic equation to a linear one, making it straightforward to solve for \(y\). Factoring reduces the complexity and helps identify or simplify solutions to polynomial equations.

Verifying Solutions

Once you've found a solution to a system of equations, it’s important to verify that the solution satisfies all original equations. This ensures that no errors were made in the solution process.
After obtaining \(y = \frac{9}{2}\) and \(x = \frac{737}{8}\), these values must be substituted back into the original equations:

• For \(x - y^3 = 1\), substituting \(x = \frac{737}{8}\) and \(y = \frac{9}{2}\) confirms \(x = y^3 + 1\).
• For \(2x = 9y^2 + 2\), substituting these values shows that the equation holds true.

Verifying ensures that all computations were accurate and confirms that the solution meets the criteria set by the system of equations. It's a vital step in solving equations that should never be skipped.

Polynomial Equations

Polynomial equations are algebraic expressions that involve variables raised to powers and their coefficients. In our problem, after substituting, we ended with a polynomial equation \(2y^3 = 9y^2\).
Polynomial equations can have various degrees, with the degree indicating the highest power of the variable. For instance, \(2y^3\) shows a cubic equation (degree 3), but we simplified it to \(2y = 9\) — a linear equation (degree 1) — through factoring.
Understanding the nature and degree of polynomial equations helps in choosing the right techniques, such as factoring or using the quadratic formula for solutions. Polynomial equations are fundamental in algebra and are often encountered in various forms, requiring techniques like the substitution and factoring we've discussed to solve effectively.