Problem 34 - Precalculus: Mathematical 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

\((x, y) = \left( \pm \sqrt[4]{\frac{69}{4}}, \sqrt[3]{\frac{1}{4}} \right) \)."

Step 1: Understand the System of Equations

We have two equations: (1) \( x^4 - y^3 = 17 \) and (2) \( 3x^4 + 5y^3 = 53 \). We are looking to find values of \( x \) and \( y \) that satisfy both equations simultaneously.

Step 2: Solve Equation (1) for y

From Equation (1) \( x^4 - y^3 = 17 \), rewrite to express \( y^3 \): \( y^3 = x^4 - 17 \).

Step 3: Substitute in Equation (2)

Substitute \( y^3 = x^4 - 17 \) into Equation (2). The equation becomes \( 3x^4 + 5(x^4 - 17) = 53 \).

Step 4: Simplify the Equation

Distribute the 5 and simplify: \[ 3x^4 + 5x^4 - 85 = 53 \].

Step 5: Combine Like Terms

Combine like terms in the equation: \( 8x^4 - 85 = 53 \).

Step 6: Solve for x

Add 85 to both sides to solve for \( x^4 \): \( 8x^4 = 138 \). Now divide by 8: \( x^4 = \frac{138}{8} \). Simplify this to get \( x^4 = \frac{69}{4} \).

Step 7: Solve for x Explicitly

Find \( x \) by taking the fourth root of \( \frac{69}{4} \): \( x = \pm \sqrt[4]{\frac{69}{4}} \).

Step 8: Solve for y Using x

Substitute \( x = \pm \sqrt[4]{\frac{69}{4}} \) back into \( y^3 = x^4 - 17 \) to find values for \( y \): since \( x^4 = \frac{69}{4} \), \( y^3 = \frac{69}{4} - 17 \). Thus, \( y^3 = \frac{69}{4} - \frac{68}{4} = \frac{1}{4} \). So \( y = \sqrt[3]{\frac{1}{4}} \).

Step 9: Combine Solutions

The solutions for the system of equations are \( (x, y) = \left( \pm \sqrt[4]{\frac{69}{4}}, \sqrt[3]{\frac{1}{4}} \right) \).

Key Concepts

Equation Solving

Solving a system of equations means finding the values of the variables that satisfy all equations involved. In this case, we have two equations that relate the variables \( x \) and \( y \). The goal is to find particular values of \( x \) and \( y \) that make both equations true simultaneously.

To solve a system, it often helps to express one variable in terms of another and then substitute this expression into other equations. Here’s what we did:

• In Equation (1): \( x^4 - y^3 = 17 \), we rearranged to solve for \( y^3 \), giving us: \( y^3 = x^4 - 17 \).
• This expression for \( y^3 \) is substituted into Equation (2) to eliminate \( y \) and find a solution in terms of \( x \).

Understanding the steps needed to isolate variables will make solving systems of equations easier. This often involves substitution and simplifying complex equations step by step.

Polynomial Equations

Polynomial equations are fundamental in algebra, involving terms like squares, cubes, and even higher powers of variables. Our problem introduces two polynomial equations, one quartic \( (x^4) \) and one cubic \( (y^3) \).

To solve these types of equations, you usually need to:

• Rearrange the equations to isolate higher power terms, such as \( x^4 \) or \( y^3 \).
• Factor or simplify where possible to reduce complexity.

Given the equations \( x^4 = \frac{69}{4} \) and \( y^3 = \frac{1}{4} \), solving involves doing arithmetic with fractional powers. It’s crucial to understand polynomial functions' properties to solve such equations efficiently.

Handling polynomial equations is a critical skill in algebra, as variably driven setups often model real-world situations.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to find solutions. This process allows you to make equations easier to handle and solve.

• In our solution, we manipulated the original equations into simpler forms. For example, we simplified \( 3x^4 + 5(x^4 - 17) = 53 \) into \( 8x^4 = 138 \) by distributing and combining like terms.
• We also turned division problems into simpler fractional equations. For example, we found \( x^4 = \frac{69}{4} \) by dividing both sides by 8.

The method of solving for \( x \) by taking the fourth root illustrates algebraic manipulation. Understanding these techniques is essential for diving deeper into more complicated algebra problems, enabling you to tackle them with confidence.