Problem 30 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} x^{4}+y^{3} &=17 \\ 3 x^{4}+5 y^{3} &=53 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((2, 1)\) and \((-2, 1)\).

Step 1: Label the equations

Label the given system of equations for easy reference. Let the first equation be Eq. (1) and the second equation be Eq. (2): \[ x^4 + y^3 = 17 \] (1) \[ 3x^4 + 5y^3 = 53 \] (2)

Step 2: Eliminate one variable

To eliminate one of the variables, multiply Eq. (1) by 3 to match the coefficients of \( x^4 \) in both equations: \[ 3(x^4 + y^3) = 3(17) \] This gives: \[ 3x^4 + 3y^3 = 51 \] (3)

Step 3: Subtract equations to solve for one variable

Subtract Eq. (3) from Eq. (2) to eliminate \( x^4 \): Eq. (2) - Eq. (3): \[ (3x^4 + 5y^3) - (3x^4 + 3y^3) = 53 - 51 \] This simplifies to: \[ 2y^3 = 2 \] Therefore, \[ y^3 = 1 \] So, \( y = 1 \) (since the cube root of 1 is 1).

Step 4: Substitute back to find the other variable

Substitute \( y = 1 \) back into Eq. (1) to find \( x \): \[ x^4 + 1^3 = 17 \] This simplifies to: \[ x^4 + 1 = 17 \] Hence, \[ x^4 = 16 \] Therefore, \( x = 2 \) or \( x = -2 \), since both 2 and -2 raised to the fourth power equal 16.

Step 5: Write final solutions for the system

Based on the values found for \( x \) and \( y \), the solutions to the system of equations are: \[ (x, y) = (2, 1) \] and \[ (x, y) = (-2, 1) \].

Key Concepts

Equation Solving

Solving equations is all about finding the values of unknown variables that satisfy given mathematical statements. When dealing with a system of equations, the goal is to find a set of values that make all the equations true simultaneously. Each equation in the system can be thought of as a condition or rule that the variables need to meet.
This process can involve a variety of techniques depending on the complexity of the equations, such as substitution, elimination, or graphical solutions.

• Identify the Equations: Clearly label each equation to avoid confusion. This makes it easier to refer back to them as you manipulate the system.
• Solve Systematically: Tackle one variable at a time and observe how changes affect others. This systematic approach is crucial for obtaining the correct result.

Equation solving is the foundation for understanding more complex mathematical concepts, as it involves careful logical reasoning and manipulation of expressions.

Variable Elimination

Variable elimination is a key technique used to reduce the complexity of a system of equations by removing one variable at a time, allowing us to solve for the others more easily. In the given exercise, variable elimination was used to simplify the process by aligning coefficients.
To eliminate a variable:

• Match Coefficients: Alter one or more equations so that the coefficients of the variable you want to eliminate are the same across compared equations. This might involve multiplying the entire equation by a constant.
• Subtract or Add Equations: Once the coefficients match, subtract one equation from the other or add them together to cancel out the variable. This step reduces the system to an equation with fewer variables, simplifying the solving process.

Variable elimination provides a clear pathway through complex systems, reducing them to more manageable parts. This technique not only simplifies calculations but also helps in focusing on one variable at a time.

Polynomial Equations

Polynomial equations are equations that involve powers of variables, such as the terms seen in the exercises: \(x^4\) and \(y^3\). Understanding these types of equations is essential, as they often appear in various mathematical challenges.

• Polynomial Forms: Polynomials can appear in many forms, but they are generally expressions with terms involving variables raised to whole-number exponents.
• Solving Polynomial Equations: The solution involves breaking down the equation into easier pieces, often by isolating terms or factoring. In our exercise, solving the equation \(x^4 = 16\) required finding the numbers that raised to the fourth power give us 16.

When working with polynomial equations, remember that the solutions may be more diverse than expected, offering both positive and negative possibilities that need to be examined. This duality is a unique feature of polynomial solutions.