Problem 83 - College Algebra with Modeling and Visualization

Question

Simplify. $$ \frac{3}{x-5}-\frac{1}{x-3}-\frac{2 x}{x-5} $$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{-2x^2 + 8x - 4}{(x-5)(x-3)}\).

Step 1: Find a Common Denominator

To combine the fractions, the denominators must be the same. The terms have denominators \((x-5)\), \((x-3)\), and \((x-5)\). The common denominator for these terms is \((x-5)(x-3)\).

Step 2: Rewrite Each Fraction

Rewrite each fraction with the common denominator: - For \(\frac{3}{x-5}\), this becomes \(\frac{3(x-3)}{(x-5)(x-3)}\).- For \(\frac{1}{x-3}\), this becomes \(\frac{1(x-5)}{(x-5)(x-3)}\).- For \(\frac{2x}{x-5}\), this becomes \(\frac{2x(x-3)}{(x-5)(x-3)}\).

Step 3: Combine the Numerators

Since all fractions now have the same denominator, combine their numerators:\[\frac{3(x-3) - 1(x-5) - 2x(x-3)}{(x-5)(x-3)}\]

Step 4: Simplify the Numerators

Distribute and combine like terms in the numerator:- Distribute to get: \[3x - 9 - (x - 5) - (2x^2 - 6x)\]- Combine terms: \(3x - 9 - x + 5 - 2x^2 + 6x = -2x^2 + 8x - 4\)

Step 5: Write Final Simplified Expression

Now the expression becomes: \[\frac{-2x^2 + 8x - 4}{(x-5)(x-3)}\] This is the final simplified form.

Key Concepts

Simplifying Fractions

Simplifying fractions is a process that makes fractions easier to read and understand. When working with algebraic fractions, it involves rewriting the fractions in a simpler form without changing their values. Algebraic fractions often contain variables in the denominators and numerators, which can make them complex at first glance. But, simplification helps in dealing with them by reducing complexity.

Here are some general steps to simplify algebraic fractions:

• Identify common factors in the numerator and the denominator.
• Cancel out common factors to reduce the fraction.
• Ensure the expression is written in the simplest form possible.

In the given exercise, the focus is on combining and simplifying terms under one common denominator. This simplifies multiple fractions into one coherent expression. Ultimately, this process doesn't just make the expression more visually appealing but also easier to work with in future calculations.

Common Denominator

Finding a common denominator is key when adding or subtracting fractions. It allows us to combine fractions that have different denominators by rewriting them to share the same one. This process ensures that the operations are performed correctly without changing the values of the fractions involved.

In this exercise, we have fractions with the denominators \(x-5\), \(x-3\), and again \(x-5\). To find a common denominator, we need to multiply these individual denominators together. This gives a common denominator of \((x-5)(x-3)\). This common multiplier is used to bring all fractions to the same base, making it possible to combine them seamlessly.

Once the fractions share a common denominator, combining them becomes as simple as adding or subtracting the numerators above this shared base. Having one denominator instead of multiple helps in carrying out these operations smoothly and efficiently.

Polynomial Division

When we talk about polynomial division in the context of algebraic fractions, we are often referring to simplifying the resulting expression after combining fractions. This isn't division in the traditional sense but rather simplification and combination techniques applied to the expression's numerator.

In the provided exercise, the operation doesn't explicitly involve division by another polynomial. However, it involves dealing with expressions in polynomial form. Post the combination of fractions under a common denominator, the numerators must be carefully handled.

Steps for polynomial simplification include:

• Distribute values across terms in parentheses.
• Carefully apply addition and subtraction among polynomials.
• Combine like terms to reach the simplest form.

After such operations, you get a simplified polynomial in the numerator, making the overall expression more concise and accessible for further algebraic manipulation, if needed. The process is about refining the expression down to its simplest, most effective form.