Problem 5 - College Algebra with Modeling and Visualization

Question

Solve the rational equation (a) symbolically, (b) graphically, and (c) numerically $$ \frac{1}{x+1}+\frac{1}{x-1}=\frac{1}{x^{2}-1} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{2}\).

Step 1: Simplify the Right Side of the Equation

First, let's simplify the right-hand side of the equation. We notice that \(x^2 - 1\) can be factored as \((x - 1)(x + 1)\). Therefore the equation becomes: \[\frac{1}{x+1} + \frac{1}{x-1} = \frac{1}{(x+1)(x-1)}\]

Step 2: Combine the Left Side into a Single Fraction

We find a common denominator for the terms on the left side, which is \((x+1)(x-1)\). Updating the left side: \[\frac{1}{x+1} = \frac{x-1}{(x+1)(x-1)} \quad \text{and} \quad \frac{1}{x-1} = \frac{x+1}{(x+1)(x-1)}\] Adding these gives: \[\frac{x-1 + x+1}{(x+1)(x-1)} = \frac{2x}{(x+1)(x-1)}\] So, the equation becomes: \[\frac{2x}{(x+1)(x-1)} = \frac{1}{(x+1)(x-1)}\]

Step 3: Eliminate the Denominators

Since both sides have the same denominator \((x+1)(x-1)\), we can safely multiply both sides by \((x+1)(x-1)\) to eliminate the denominators: \[2x = 1\]

Step 4: Solve for x

Now, solve the simple equation: \[2x = 1\] To find \(x\), divide both sides by 2: \[x = \frac{1}{2}\]

Step 5: Confirm No Division by Zero

Check that the solution \(x = \frac{1}{2}\) does not make any denominator zero in the original equation. Substitute \(x = \frac{1}{2}\) into denominators: \((x + 1) = \frac{3}{2}\), \((x - 1) = -\frac{1}{2}\), \((x^2 - 1) eq 0\) at \(x = \frac{1}{2}\). Thus, it is a valid solution.

Step 6: Graphical Solution

The function \(f(x) = \frac{1}{x+1} + \frac{1}{x-1}\) and \(g(x) = \frac{1}{x^2-1}\) are plotted. The intersection point(s) of these graphs gives the solution. At \(x = \frac{1}{2}\), these two functions intersect, confirming our solution found symbolically.

Step 7: Numerical Solution via Substitution

Substitute \(x = \frac{1}{2}\) into the original equation: \(\frac{1}{\frac{1}{2}+1} + \frac{1}{\frac{1}{2}-1} = \frac{1}{(\frac{1}{2})^2-1}\). Both sides calculate to \(-4\), confirming that \(x = \frac{1}{2}\) satisfies the equation.

Key Concepts

Factoring Polynomials

Factorization is a crucial skill when dealing with algebraic expressions, especially rational equations. In the given problem, the right side involves the expression \(x^2 - 1\). Recognizing this as a difference of squares allows us to factor it into \((x-1)(x+1)\). This factorization simplifies solving the equation, as it allows us to match the denominators on both sides.

Factoring helps to:

• Simplify complex expressions.
• Identify common denominators when dealing with fractions.
• Reveal possible values that can lead to zero denominators, avoiding undefined expressions.
  
  

It’s essential to factor polynomials effectively in rational equations to simplify and accurately solve them. Always check the factored form provides all solutions or excludes problematic values.

Solving Equations

Solving equations like our rational equation involves several steps, which need to be clearly understood. First, simplify both sides of the equation as much as possible. Then, combine and create a single fraction with a common denominator for terms on either side. In our case, both sides have the denominator \((x+1)(x-1)\).

To solve the equation:

• Simplify the original expressions.
• Combine fractions on each side to have a common denominator.
• Eliminate denominators by multiplying through, turning the equation into a simpler linear equation when possible.
• Solve the resulting equation; for instance, \(2x = 1\) leads us to \(x = \frac{1}{2}\).

Always verify that solutions don’t result in zero denominators, as these values are excluded from the solution set.

Graphical Methods

Graphical methods offer a visual approach to solving equations. By plotting the functions representing each side of the equation, we can find where the graphs intersect. This intersection point provides the solution to the equation.

To use graphical methods effectively:

• Plot each side of the equation as a separate function. For our example, set \(f(x)\) to \(\frac{1}{x+1} + \frac{1}{x-1}\) and \(g(x)\) to \(\frac{1}{x^2-1}\).
• Identify intersection points, which indicate the solution to the equation.
• Ensure the intersection is at a valid domain for both functions, not where any function is undefined.
  
  

This approach is especially helpful in complex equations, offering an intuitive check against symbolic solutions.

Numerical Methods

Numerical methods can verify solutions found through other approaches, such as solving or graphing. Although numerical methods aren't always needed for straightforward problems, they serve as an excellent verification technique.

In our exercise, after solving \(2x = 1\), confirming \(x = \frac{1}{2}\) is a solution involves substituting back into the original equation:

• Compute \(\frac{1}{\frac{1}{2}+1} + \frac{1}{\frac{1}{2}-1} \). Simplifies and evaluates both sides.
• Ensure the calculated values align on both sides of the equation.
• Double-check these computations to confirm consistency and reliability of results.
  
  

Numerical verification is a strong tool to ensure solutions are accurate and consistent with initial symbolic solving.