Problem 60 - Elementary and Intermediate Algebra

Question

Solve for the specified variable or expression. $$ b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2} \text { for } b^{2} $$

Step-by-Step Solution

Answer

\( b^2 = \frac{a^2 y^2 x^2}{a^2 - x^2} \).

Step 1: Isolate the Term Involving \( b^2 \)

To solve the equation \( b^2 x^2 + a^2 y^2 = a^2 b^2 \) for \( b^2 \), we first aim to isolate the term that contains \( b^2 \). Begin by subtracting \( a^2 y^2 \) from both sides, leading to the equation \( b^2 x^2 = a^2 b^2 - a^2 y^2 \).

Step 2: Factor the Right Side of the Equation

Notice that the right side, \( a^2 b^2 - a^2 y^2 \), can be factored. Extract \( a^2 \) as a common factor, which yields: \( b^2 x^2 = a^2 (b^2 - y^2) \).

Step 3: Divide Both Sides by \( x^2 \)

Divide both sides of the equation \( b^2 x^2 = a^2 (b^2 - y^2) \) by \( x^2 \) to solve for \( b^2 \). This simplifies to \( b^2 = \frac{a^2 (b^2 - y^2)}{x^2} \).

Step 4: Simplify and Solve for \( b^2 \)

Rearrange to solve explicitly for \( b^2 \). We get \( b^2 = \frac{a^2 b^2}{x^2} - \frac{a^2 y^2}{x^2} \). Isolate \( b^2 \): \( b^2 \left(1 - \frac{a^2}{x^2}\right) = -\frac{a^2 y^2}{x^2} \). Simplify to \( b^2 = \frac{a^2 y^2}{\frac{a^2}{x^2} - 1} \).

Step 5: Simplify the Expression

Further simplify the expression: factoring further, we have \( b^2 = \frac{a^2 y^2 x^2}{a^2 - x^2} \). Conclude the simplification: \( b^2 = x^2y^2 \) (assuming correct setup and simplifications that lead to rational function resolution).

Key Concepts

Solving Equations

Solving equations involves finding the value of one or more variables that satisfy a given equation. It often starts with isolating the variable, which means getting the variable on one side of the equation by itself.

In this case, we're tasked with solving for \( b^2 \) in the equation \( b^2 x^2 + a^2 y^2 = a^2 b^2 \). The first step includes subtracting \( a^2 y^2 \) from both sides to isolate \( b^2 x^2 \) on one side.

After this initial move, the rest of the process involves continuing to manipulate the equation algebraically until \( b^2 \) is by itself on one side of the equation. Each step follows logically from the previous step to systematically approach the solution.

Algebraic Manipulation

Algebraic manipulation refers to using mathematical operations and properties to change the form of an equation, without changing its meaning. This is a vital skill to solve equations like the one we have.

First, we manipulate the equation by factoring out common terms, as seen with \( a^2 \) in the equation \( b^2 x^2 = a^2 (b^2 - y^2) \).

Another manipulation involves dividing both sides of an equation by a variable or a number to simplify it further. These strategic moves allow us to see patterns or structures in the equation that facilitate finding a solution.

Factoring Expressions

Factoring is the process of breaking down an expression into simpler elements, often to make solving an equation easier. It’s like turning a complex material into manageable pieces.

In our step-by-step solution, we factor the right side of the equation by identifying \( a^2 \) as a common factor in \( a^2 b^2 - a^2 y^2 \).

Once factored, the expression becomes \( a^2 (b^2 - y^2) \), which makes it easier to manage and manipulate in subsequent steps to solve for \( b^2 \). Recognizing these kinds of opportunities for factoring is critical to simplify complex algebraic problems effectively.

Rational Functions

Rational functions are ratios of polynomials. They're an important concept in algebra, especially when manipulating complex expressions.

In this problem, the final solution for \( b^2 \) becomes a rational expression: \( b^2 = \frac{a^2 y^2 x^2}{a^2 - x^2} \).

This form reveals how the value of \( b^2 \) is expressed in terms of other variables and constants, highlighting the interconnectedness of the equation's components. Rational functions allow us to see how changes in certain variables affect the overall expression. This understanding is crucial for analyzing and solving advanced algebraic problems efficiently.