Problem 92 - College Algebra with Modeling and Visualization

Question

Simplify. $$ \frac{3}{x^{2}-x-6}-\frac{2}{x^{2}+5 x+6} $$

Step-by-Step Solution

Answer

\( \frac{x + 15}{(x-3)(x+2)(x+3)} \)

Step 1: Factor the Denominators

First, factor the quadratic expressions in the denominators. For the first fraction, \(x^2 - x - 6\) can be factored into \((x-3)(x+2)\). For the second fraction, \(x^2 + 5x + 6\) can be factored into \((x+2)(x+3)\).

Step 2: Determine the Least Common Denominator (LCD)

For the least common denominator, take each factor from the factorizations of the denominators without repetition. Thus, the LCD is \((x-3)(x+2)(x+3)\).

Step 3: Express Each Fraction with the LCD

Rewrite each fraction such that the denominators are the LCD \((x-3)(x+2)(x+3)\). Multiply the first fraction by \(\frac{x+3}{x+3}\) and the second fraction by \(\frac{x-3}{x-3}\), so:\[ \frac{3(x+3)}{(x-3)(x+2)(x+3)} - \frac{2(x-3)}{(x-3)(x+2)(x+3)} \]

Step 4: Subtract the Numerators

Since the denominators are now the same, subtract the numerators:\[ 3(x+3) - 2(x-3) \]Distribute the constants:\[ 3x + 9 - 2x + 6 \]Combine like terms to get:\[ x + 15 \]

Step 5: Write the Simplified Expression

Combine the results into a single fraction:\[ \frac{x + 15}{(x-3)(x+2)(x+3)} \]

Key Concepts

Factoring Quadratics

Understanding how to factor quadratics is key when working with rational expressions. Factoring is the process of breaking down a quadratic expression into a product of simpler expressions. For example, the expression \(x^2 - x - 6\) can be factored. We aim for two numbers that multiply to \(-6\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)). These numbers are \(-3\) and \(+2\), resulting in the factors \((x-3)(x+2)\). Similarly, for \(x^2 + 5x + 6\), find two numbers that multiply to \(6\) and add to \(5\), which are \(+2\) and \(+3\), leading to \((x+2)(x+3)\). Factoring quadratics makes it easier to manipulate expressions, especially when looking for the least common denominator in fractions.

Least Common Denominator

The least common denominator (LCD) is crucial when adding or subtracting fractions with different denominators. The LCD is the smallest expression or product of expressions that each denominator can be divided by without a remainder. It allows us to combine the fractions efficiently.To find the LCD for the rational expressions involved, take each distinct factor from the denominators while considering the highest power of each factor. For the fractions \(\frac{3}{(x-3)(x+2)}\) and \(\frac{2}{(x+2)(x+3)}\), the LCD is \((x-3)(x+2)(x+3)\). Using the LCD simplifies subtraction or addition of fractions by converting them to have the same denominator.

Subtracting Fractions

Once you've factored the quadratics and found the least common denominator, the next step is subtracting fractions. When fractions have the same denominator, you can subtract the numerators directly.Given: \[ \frac{3(x+3)}{(x-3)(x+2)(x+3)} - \frac{2(x-3)}{(x-3)(x+2)(x+3)} \]First, subtract the expressions in the numerators:- Distribute: \(3(x+3)\) gives \(3x + 9\), and \(-2(x-3)\) gives \(-2x + 6\).- Simplify: \(3x + 9 - 2x + 6 = x + 15\).Finally, the fraction \(\frac{x + 15}{(x-3)(x+2)(x+3)}\) is the simplified result of the subtraction. It's essential to always check if further simplification is possible.