Problem 35 - Precalculus: Functions and Graphs

Question

Simplify the expression. $$\frac{\frac{x}{x+2}-\frac{4}{x+2}}{x-3-\frac{6}{x+2}}$$

Step-by-Step Solution

Answer

\( \frac{x-4}{(x-3)(x+2)} \)

Step 1: Simplify the Numerator

The numerator of the given expression is \( \frac{x}{x+2} - \frac{4}{x+2} \). Since both fractions have the same denominator, we can directly subtract the numerators: \( x - 4 \). Hence, the simplified form becomes \( \frac{x-4}{x+2} \).

Step 2: Simplify the Denominator

The denominator is \( x - 3 - \frac{6}{x+2} \). To write this with a common denominator, express \( x - 3 \) as \( \frac{(x-3)(x+2)}{x+2} \). This becomes \( \frac{x^2 + 2x - 3x - 6}{x+2} - \frac{6}{x+2} = \frac{x^2 - x - 6}{x+2} \).

Step 3: Substitute Simplified Parts and Simplify Entire Expression

Now substitute the simplified numerator and denominator back into the original expression: \( \frac{\frac{x-4}{x+2}}{\frac{x^2 - x - 6}{x+2}} \). This simplifies to \( \frac{x-4}{x^2 - x - 6} \) since the \( x+2 \) terms cancel out.

Step 4: Factor the Denominator

Factor the quadratic expression in the denominator: \( x^2 - x - 6 = (x-3)(x+2) \). This enables further simplification.

Step 5: Simplify Final Expression

Substitute the factored form back to simplify: \( \frac{x-4}{(x-3)(x+2)} \). Cross-check if the numerator shares common factors with the denominator. Since there are no common factors, this is the simplest form.

Key Concepts

Factoring Quadratics

In mathematics, quadratics are equations of the form \( ax^2 + bx + c \). These equations can often be simplified by factoring, which involves breaking down the quadratic into simpler expressions that, when multiplied together, yield the original equation. In our exercise, the quadratic expression \( x^2 - x - 6 \) is factored into \((x-3)(x+2)\). This step is crucial for simplification, as it allows us to see potential cancellations in a rational expression.

Here's a quick way to factor:

• Identify two numbers that multiply into \( c \) (the constant term) and add up to \( b \) (the coefficient of \( x \)).
• In this case, we need numbers that multiply to \(-6\) and add to \(-1\). These numbers are \( -3 \) and \( 2 \).
• Thus, \( x^2 - x - 6 \) factors into \((x-3)(x+2)\).

This approach works efficiently for simpler quadratics and is essential for understanding rational expressions.

Common Denominators

When working with fractions, finding a common denominator is a crucial step in simplifying expressions. The common denominator allows us to combine or subtract fractions by ensuring they share the same baseline. In the given exercise, both the numerator and denominator in the expression required finding common denominators. Without this step, it would be impossible to merge the fractions together.

Here’s how it works:

• Take the original fractions—if they already have the same denominator, like in the expression \( \frac{x}{x+2} - \frac{4}{x+2} \), you can immediately subtract the numerators, resulting in \( \frac{x-4}{x+2} \).
• If they have different denominators, as in the denominator of the main fraction from the exercise, you’ll adjust one or both fractions to have the same denominator.
• Once everything is under a common denominator, operations become straightforward, allowing for simplification steps.

Using common denominators is a foundation skill necessary in algebra, especially when dealing with more complex algebraic fractions.

Algebraic Fractions

Algebraic fractions are similar to numerical fractions, but they consist of polynomials in the numerator or denominator. Just like simpler fractions, they can often be simplified by canceling common factors. In our problem, the complexity of algebraic fractions arises because these include variables and require factoring and distribution to simplify.

Consider the expression \( \frac{x-4}{(x-3)(x+2)} \). Here's the approach:

• First, simplify both the numerator and the denominator by canceling any common factors, if they exist.
• In our solution, the factored form \((x-3)(x+2)\) is left intact because no common factors appear with \(x-4\).
• Finally, verify the simplified form for any restrictions that occur due to denominators potentially being zero.

Understanding algebraic fractions involves grasping both numerical and algebraic properties, making it essential to master basic operations of addition, subtraction, and division of polynomials.

Precalculus

Precalculus lays the groundwork for calculus, blending previously established concepts from algebra and trigonometry. It provides the mathematical methodologies that underpin calculus, especially those involving function behavior, graphing, and advanced algebraic manipulation.

In the context of simplifying rational expressions—an integral part of precalculus—students enhance their technical skills, learning to manipulate expressions into more workable forms.

• Algebraic simplification techniques, like those demonstrated in our exercise, streamline complex rational expressions into understandable terms.
• Factoring, as used to break down quadratics, is a particular focus in precalculus to prepare for more sophisticated calculus operations.
• This preparation helps students tackle calculus concepts, like limits and continuity, that often require simplified algebraic forms.

By engaging with this kind of algebraic manipulation early, students build the confidence and competence needed to succeed in calculus.