Problem 32 - Precalculus: Mathematics 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 \((\pm\sqrt{2}, \pm\sqrt{2})\).

Step 1: Introduce Substitutions

To make the equations simpler, let's set \( a = \frac{1}{x^2} \) and \( b = \frac{1}{y^4} \). This reduces the system to: \( 4a + 6b = \frac{7}{2} \) and \( a - 2b = 0 \).

Step 2: Solve the Second Equation for a

From the equation \( a - 2b = 0 \), we can solve for \( a \) in terms of \( b \), giving us \( a = 2b \).

Step 3: Substitute a into the First Equation

Substitute \( a = 2b \) into the first equation: \( 4(2b) + 6b = \frac{7}{2} \). Simplifying gives \( 8b + 6b = \frac{7}{2}\).

Step 4: Solve for b

Combine like terms in the equation: \( 14b = \frac{7}{2} \). Divide both sides by 14 to solve for \( b \): \( b = \frac{1}{4} \).

Step 5: Find a Using b

Since \( a = 2b \) and \( b = \frac{1}{4} \), substituting gives \( a = 2 \times \frac{1}{4} = \frac{1}{2} \).

Step 6: Solve for x and y

Recalling our earlier substitutions, \( a = \frac{1}{x^2} \) implies \( \frac{1}{x^2} = \frac{1}{2} \), so \( x^2 = 2 \), giving solutions \( x = \pm \sqrt{2} \). For \( b = \frac{1}{y^4} = \frac{1}{4} \), \( y^4 = 4 \), leading to \( y = \pm \sqrt{2} \).

Step 7: Determine the Solution Pairs

The solutions for \( x \) are \( \pm \sqrt{2} \) and for \( y \) are \( \pm \sqrt{2} \). Combine these to form the solution pairs: \( (x,y) = (\sqrt{2}, \sqrt{2}), (\sqrt{2}, -\sqrt{2}), (-\sqrt{2}, \sqrt{2}), (-\sqrt{2}, -\sqrt{2}) \).

Key Concepts

Substitution Method

The substitution method is a handy tool for tackling systems of equations, especially when they involve complex terms. It involves expressing one variable in terms of another and then substituting this expression into another equation. This simplifies the system and helps isolate each variable.
To see how this works, consider our initial system of rational equations:

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

We introduced substitutions by letting \( a = \frac{1}{x^2} \) and \( b = \frac{1}{y^4} \). This allowed us to transform the system into simpler linear equations:

• \( 4a + 6b = \frac{7}{2} \)
• \( a - 2b = 0 \)

By expressing \( a \) in terms of \( b \) from the second equation, and substituting back into the first equation, the complexity is greatly reduced. This method is particularly useful when either equation can be easily transformed or when dealing with algebraic expressions like powers or fractions.

Solution Pairs

When solving systems of equations, identifying solution pairs refers to finding all possible pairs of values for the variables that satisfy all the equations simultaneously. This is key when dealing with multiple equations that intersect, as different intersections can define different solutions.
In the original exercise, after calculating the values for \( a \) and \( b \), we found potential solutions for \( x \) and \( y \):

• For \( x \), solutions were \( x = \pm \sqrt{2} \)
• For \( y \), solutions were \( y = \pm \sqrt{2} \)

Each combination of these values forms a valid solution pair - meaning, any of these pairs solves the original system entirely. Thus, we identify four solution pairs:

• \((\sqrt{2}, \sqrt{2})\)
• \((\sqrt{2}, -\sqrt{2})\)
• \((-\sqrt{2}, \sqrt{2})\)
• \((-\sqrt{2}, -\sqrt{2})\)

These solutions illustrate how systems of equations can have more than one set of solution pairs.

Rational Equations

Rational equations involve fractions with expressions in their denominators. These equations require particular strategies for solving, given their complex nature. They involve variables in the denominator, hence a high degree of care is needed to avoid errors, like division by zero.
In our problem, two rational equations were given:

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

To simplify solving these rational equations, substitutions were made to turn them into linear equations. This approach mitigates direct manipulation with fractions and powers, which could otherwise become cumbersome.
Moreover, when these equations are solved, the solutions must be verified back in the original equations to ensure they do not make any denominators zero, which could invalidate the results. Therefore, rational equations call for thorough verification of solution pairs, ensuring every step follows mathematically sound principles.