Problem 36 - Precalculus: Mathematical for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l}\frac{4}{x^{2}}+\frac{6}{y^{4}}=\frac{7}{2} \\\\\frac{1}{x^{2}}-\frac{2}{y^{4}}=0\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{2},\sqrt{2}), (-\sqrt{2},\sqrt{2}), (\sqrt{2},-\sqrt{2}), (-\sqrt{2},-\sqrt{2})\).

Step 1: Understand the System of Equations

We have two equations: 1) \( \frac{4}{x^{2}} + \frac{6}{y^{4}} = \frac{7}{2} \) 2) \( \frac{1}{x^{2}} - \frac{2}{y^{4}} = 0 \). Our goal is to find values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solve the Second Equation for \( \frac{1}{x^{2}} \)

From the second equation \( \frac{1}{x^{2}} - \frac{2}{y^{4}} = 0 \), we can solve for \( \frac{1}{x^{2}} \):\[ \frac{1}{x^{2}} = \frac{2}{y^{4}} \].This relation will be used in the first equation.

Step 3: Substitute in the First Equation

Using the expression from Step 2, substitute \( \frac{1}{x^{2}} = \frac{2}{y^{4}} \) into the first equation:\[ 4 \cdot \left( \frac{2}{y^{4}} \right) + \frac{6}{y^{4}} = \frac{7}{2} \].Simplify it to:\[ \frac{8}{y^{4}} + \frac{6}{y^{4}} = \frac{7}{2} \].

Step 4: Simplify the Equation

Add the fractions on the left side:\[ \frac{8 + 6}{y^{4}} = \frac{7}{2} \]which simplifies to:\[ \frac{14}{y^{4}} = \frac{7}{2} \].

Step 5: Solve for \( y^4 \)

Cross multiply to eliminate the fractions:\[ 14 \cdot 2 = 7 \cdot y^{4} \]gives us:\[ 28 = 7y^{4} \].Divide both sides by 7 to find \( y^4 \):\[ y^{4} = 4 \].

Step 6: Solve for \( y \)

We have \( y^4 = 4 \). Solve for \( y \):\[ y = \pm \sqrt[4]{4} \].Since \( \sqrt[4]{4} = \sqrt{2} \), the solutions are:\( y = \pm \sqrt{2} \).

Step 7: Find Corresponding \( x \) Values

Use \( \frac{1}{x^{2}} = \frac{2}{y^{4}} \) to find \( x \):For \( y = \sqrt{2} \) or \( y = -\sqrt{2} \), substitute \( y^{4} = 4 \) into \( \frac{1}{x^{2}} = \frac{2}{4} \):\[ \frac{1}{x^{2}} = \frac{1}{2} \].This gives \( x^2 = 2 \), so \( x = \pm \sqrt{2} \).

Step 8: List All Solution Pairs

Combine the values for \( x \) and \( y \) from Steps 6 and 7:- \( (x, y) = (\sqrt{2}, \sqrt{2}) \)- \( (x, y) = (-\sqrt{2}, \sqrt{2}) \)- \( (x, y) = (\sqrt{2}, -\sqrt{2}) \)- \( (x, y) = (-\sqrt{2}, -\sqrt{2}) \).These are all the solutions that satisfy both equations.

Key Concepts

Solving Equations

Understanding how to solve a system of equations is a crucial concept in mathematics. A system of equations consists of two or more equations with the same set of variables. The task is to find values for these variables that satisfy all equations in the system simultaneously. In this exercise, we were given two equations involving the variables \( x \) and \( y \) in rational forms:

• \( \frac{4}{x^{2}} + \frac{6}{y^{4}} = \frac{7}{2} \)
• \( \frac{1}{x^{2}} - \frac{2}{y^{4}} = 0 \)

To find the solution, we must manipulate these relationships to express one variable in terms of the other, or simplify them to find common values of \( x \) and \( y \). This process requires several algebraic techniques, including substitution, to solve the equations step by step.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, where we solve one of the equations for one variable and then substitute this result into the other equations. In this exercise, we began by solving the second equation for \( \frac{1}{x^{2}} \):

• Start with \( \frac{1}{x^{2}} - \frac{2}{y^{4}} = 0 \).
• Add \( \frac{2}{y^{4}} \) to both sides to isolate \( \frac{1}{x^{2}} \), resulting in \( \frac{1}{x^{2}} = \frac{2}{y^{4}} \).

This result becomes particularly useful as the expression \( \frac{1}{x^{2}} \) can now be used in place of \( x \) in the first equation. By substituting \( \frac{1}{x^{2}} = \frac{2}{y^{4}} \) into our first equation, the problem transforms into a single-variable equation in terms of \( y \). This simplifies the process considerably and allows us to solve for \( y \) before back-substituting to find \( x \).

Algebraic Manipulation

Algebraic manipulation is essential when working with systems of equations. It involves rearranging terms, factoring, and simplifying expressions to make them more manageable. In our example, after substituting for \( \frac{1}{x^{2}} \), we focused on the first equation:

• Substitute to get: \( \frac{8}{y^{4}} + \frac{6}{y^{4}} = \frac{7}{2} \).
• Combine terms to get: \( \frac{14}{y^{4}} = \frac{7}{2} \).

To remove fractions, cross-multiply, which eradicates denominators and allows solving for \( y^4 \):

• Cross multiply to find \( 14 \cdot 2 = 7 \cdot y^{4} \).
• Solve for \( y^{4} \) yielding: \( y^{4} = 4 \).

Completing the solution for \( y \) involved taking the fourth root, giving us \( y = \pm \sqrt[4]{4} = \pm \sqrt{2} \). With \( y \) known, we refocus on solving for \( x \) using the relationship \( \frac{1}{x^{2}} = \frac{1}{2} \), which translates to \( x = \pm \sqrt{2} \). This structured approach exemplifies the nature of algebraic manipulation in solving equations effectively.