Problem 22 - College Algebra with Modeling and Visualization

Question

Find all real solutions. Check your results. $$ \frac{1}{x-1}+\frac{3}{x+1}=\frac{4}{x^{2}-1} $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{3}{2} \).

Step 1: Simplify the Equation

Notice that the expression on the right side, \( \frac{4}{x^2 - 1} \), can be rewritten. Since \( x^2 - 1 = (x-1)(x+1) \), rewrite the equation as: \( \frac{1}{x-1} + \frac{3}{x+1} = \frac{4}{(x-1)(x+1)} \).

Step 2: Find a Common Denominator

The left-hand side of the equation has fractions with denominators \( x-1 \) and \( x+1 \). The common denominator is \( (x-1)(x+1) \). Express each term with this common denominator: \( \frac{(x+1)}{(x-1)(x+1)} + \frac{3(x-1)}{(x-1)(x+1)} = \frac{4}{(x-1)(x+1)} \).

Step 3: Combine and Simplify

Combine the terms on the left: \( \frac{x+1 + 3(x-1)}{(x-1)(x+1)} = \frac{x+1 + 3x - 3}{(x-1)(x+1)} \). Simplify the numerator: \( \frac{4x - 2}{(x-1)(x+1)} \). Now the equation is \( \frac{4x - 2}{(x-1)(x+1)} = \frac{4}{(x-1)(x+1)} \).

Step 4: Solve the Numerator Equation

Since the denominators are the same, equate the numerators: \( 4x - 2 = 4 \). Solve for \( x \): \( 4x = 6 \), so \( x = \frac{3}{2} \).

Step 5: Check for Extraneous Solutions

Verify whether \( x = \frac{3}{2} \) is valid by ensuring it doesn't invalidate any of the original denominators. Substitute back into the denominators \( x-1 \) and \( x+1 \): \( x-1 = \frac{1}{2} \) and \( x+1 = \frac{5}{2} \) are non-zero. Hence, \( x = \frac{3}{2} \) is a valid solution.

Key Concepts

Rational Equations

Rational equations involve fractions that include variables in their denominators. Such equations can seem complex at first, but with the right steps, they become manageable.When faced with a rational equation, the first step is to simplify each fraction. This involves rewriting complex expressions. In our example, the expression \( \frac{4}{x^2 - 1} \) becomes \( \frac{4}{(x-1)(x+1)} \). This is possible by identifying that \( x^2 - 1 \) can be factored into \((x-1)(x+1)\).Rational equations often require combining fractions, which sets the stage for solving the equation. Just remember that whatever operations you perform on fractions, respecting the rules of arithmetic and algebra is crucial.

Common Denominator

A crucial concept in handling rational equations is finding a common denominator. This is necessary when you need to combine fractions that do not initially share the same base.For our equation, the denominators \( x-1 \) and \( x+1 \) need a common denominator to combine effectively. Multiplying these gives \((x-1)(x+1)\), which unifies all fractions involved.Once each fraction is expressed with this common denominator, as seen in the step \( \frac{(x+1)}{(x-1)(x+1)} + \frac{3(x-1)}{(x-1)(x+1)} \), simplifying is much more straightforward. Having the same denominator means you can directly sum the numerators.

Real Solutions

The objective of solving rational equations is to find all real solutions. A real solution is a value of \( x \) that satisfies the equation, yielding valid results when substituted back into the original fractions.In the example provided, solving \( 4x - 2 = 4 \) gives \( x = \frac{3}{2} \). However, it's important to verify that this solution is valid for the original equation. Checking for any x that would make any denominator zero avoids extraneous solutions. Since \( x = \frac{3}{2} \) does not make \( x-1 \) or \( x+1 \) zero, it is valid.Always substitute potential solutions back into the original context to ensure they don't break any mathematical rules, especially regarding rational expressions.