Problem 13 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation by hand. $$\frac{2 x}{x^{2}-1}=\frac{2}{x+1}-\frac{1}{x-1}$$

Step-by-Step Solution

Answer

The solution is \(x = -3\).

Step 1: Find a common denominator

To solve the equation \(\frac{2 x}{x^{2}-1}=\frac{2}{x+1}-\frac{1}{x-1}\), we first recognize that \(x^2 - 1\) can be factored as \((x-1)(x+1)\). Therefore, the common denominator for the entire equation is \((x-1)(x+1)\). We will use this to rewrite each term under a common denominator.

Step 2: Rewrite the equation using the common denominator

Rewrite each fraction using the common denominator:- The left side becomes \(\frac{2x}{(x-1)(x+1)}\).- The right side's first fraction is \(\frac{2(x-1)}{(x-1)(x+1)}\).- The right side's second fraction is \(\frac{1(x+1)}{(x-1)(x+1)}\).Thus, the equation becomes: \[ \frac{2x}{(x-1)(x+1)} = \frac{2(x-1)}{(x-1)(x+1)} - \frac{1(x+1)}{(x-1)(x+1)} \].

Step 3: Combine the right side

Combine the fractions on the right side over the common denominator:\[ \frac{2(x-1) - (x+1)}{(x-1)(x+1)} = \frac{2x - 2 - x - 1}{(x-1)(x+1)} = \frac{x - 3}{(x-1)(x+1)} \].Now the equation is:\[ \frac{2x}{(x-1)(x+1)} = \frac{x - 3}{(x-1)(x+1)} \].

Step 4: Clear the denominators

Since both sides of the equation now have the same denominator, we can set the numerators equal to each other:\(2x = x - 3\).

Step 5: Solve for x

Isolate \(x\) by subtracting \(x\) from both sides:\(2x - x = -3\)This simplifies to:\(x = -3\).

Step 6: Verify the solution

Substitute \(x = -3\) back into the original equation to check for any extraneous solutions:- Left side: \(\frac{2(-3)}{(-3)^2-1} = \frac{-6}{8}\).- Right side: \(\frac{2}{-2} - \frac{1}{-4} = -1 + \frac{1}{4} = -\frac{3}{4}\).Both sides equal \(-\frac{3}{4}\), confirming \(x = -3\) is a valid solution.

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial step in simplifying complex equations. When you encounter a polynomial such as \(x^2 - 1\), it can often be rewritten to make solving the equation easier.
In this case, \(x^2 - 1\) can be factored into \((x-1)(x+1)\).

• Factoring helps in identifying common roots and expressing the polynomial in a more manageable form.
• This technique simplifies operations like addition, subtraction, or finding common denominators of algebraic fractions.

Grasping this concept is essential since it often forms the backbone of many algebraic solutions.

Common Denominator

Finding a common denominator is a pivotal step in solving equations that involve fractions. In the given equation, \(\frac{2 x}{x^{2}-1} = \frac{2}{x+1} - \frac{1}{x-1}\), each side has different denominators initially.
To solve, you need to convert these into a common denominator to manage the fractions effectively.

• The common denominator here is \((x-1)(x+1)\), which simplifies the equation handling.
• Doing so ensures that each fraction can be combined or compared easily.

This simplification is fundamental because matched denominators let us equate or combine fractions in a unified form.

Cross-Multiplication

Once the fractions have a common denominator on both sides of an equation, you can effectively clear these denominators using cross-multiplication. Here, it involves setting the numerators equal since the denominators are already the same.
This method transforms \(\frac{2x}{(x-1)(x+1)} = \frac{x - 3}{(x-1)(x+1)}\) into the simpler form of equation, \(2x = x - 3\).

• This step is pivotal because it converts a complex fraction into a straightforward linear equation.
• With the numerators isolated, solving becomes a matter of basic algebra, enhancing clarity and focus on solving for the variable.

Understanding this approach helps in tackling a broad array of algebraic problems more efficiently.

Verification of Solutions

After calculating potential solutions, verifying them is key to ensuring accuracy. When \(x = -3\) was found as a solution, it was necessary to substitute it back into the original equation to confirm its validity.
Here's how:

• Substituting \(x = -3\) into both sides yielded \(-\frac{3}{4}\) as the result, confirming the solution.
• This verification step checks for any errors in computation or extraneous solutions that may not satisfy the original equation.

Such verification is a vital part of problem-solving, providing a reliable conclusion to your algebraic journey.