Problem 26 - College Algebra with Modeling and Visualization

Question

Find all real solutions. Check your results. $$ \frac{1}{x-3}+1=\frac{6}{x^{2}-9} $$

Step-by-Step Solution

Answer

The only real solution is \(x = -4\).

Step 1: Identify and Simplify Expressions

The given equation is \( \frac{1}{x-3}+1=\frac{6}{x^{2}-9} \). Notice that \( x^2 - 9 \) can be factored as \((x-3)(x+3)\). Rewriting the equation, we have: \( \frac{1}{x-3}+1=\frac{6}{(x-3)(x+3)} \).

Step 2: Eliminate the Fractions

Clear the fractions by multiplying every term by \((x-3)(x+3)\), recognizing that \(x eq 3\) and \(x eq -3\) to avoid division by zero. This gives: \[(x + 3) + (x-3)(x+3) = 6\].

Step 3: Simplify the Equation

Notice that \((x-3)(x+3) = x^2 - 9 \). Substituting back gives:\[x + 3 + x^2 - 9 = 6\] Simplifying further, this becomes:\[x^2 + x - 6 = 6\].

Step 4: Solve the Quadratic Equation

Subtract 6 from both sides to set the equation to zero: \[x^2 + x - 12 = 0\]. Now solve the quadratic equation. We can factor this as \((x - 3)(x + 4) = 0\).

Step 5: Find the Real Solutions

Setting each factor equal to zero gives \(x - 3 = 0\) or \(x + 4 = 0\). Solving these, we find \(x = 3\) and \(x = -4\).

Step 6: Check for Valid Solutions

Substitute \(x = 3\) back into the original equation and note it makes the denominator zero, so \(x = 3\) is not valid. Substitute \(x = -4\): \[\frac{1}{-4 - 3} + 1 = \frac{6}{16 - 9}\] which simplifies to \[ -\frac{1}{7} + 1 = \frac{6}{7}\], holding true. Therefore, \(x = -4\) is a valid solution.

Key Concepts

Factoring Polynomials

Factoring polynomials is a skill that helps simplify complex mathematical expressions. In our exercise, we encountered the expression \( x^2 - 9 \). This is a special polynomial known as a "difference of squares". It can be factored into two binomials:

• \( x^2 - 9 = (x-3)(x+3) \)

These binomials

1. \( (x-3) \)
2. \( (x+3) \)

help simplify the equation by reducing its complexity. Recognizing such patterns is key to factoring efficiently. As a result, the equation can be rewritten in a simpler form, making subsequent steps more manageable.

Eliminating Fractions in Equations

Fractions can make equations seem complex, but they can be eliminated to simplify calculation. In the given equation, we have fractions such as \( \frac{1}{x-3} \) and \( \frac{6}{(x-3)(x+3)} \). The goal is to eliminate these by clearing the denominators. Here's how:

• Identify the least common denominator (LCD), which in this case is \( (x-3)(x+3) \).
• Multiply each term by this LCD to clear fractions, transforming the equation into a polynomial form.

This results in

1. \((x + 3) + (x-3)(x+3) = 6\)

After these steps, the equation becomes polynomial, allowing for easier manipulation and solving.

Checking Solutions in Algebra

After finding solutions to an algebra problem, it is crucial to verify them. This process ensures the accuracy and validity of your results. Let's break down how to check solutions:

• Substitute each solution back into the original equation.
• Verify if both sides of the equation equal each other.

In our example, we found solutions \(x = 3\) and \(x = -4\). When substituting \(x = 3\) into the original equation, it results in division by zero, making it invalid. However, substituting \(x = -4\) fits perfectly:

1. The left side simplifies to \( -\frac{1}{7} + 1 \)
2. The right evaluates to \( \frac{6}{7} \)

Both sides match, confirming \(x = -4\) as the valid solution. Checking solutions helps prevent mathematical errors and ensures the reliability of your answers.