Problem 30 - Algebra for College Students

Question

Perform the indicated operations, and express your answers in simplest form. $$ \frac{2 x-1}{x+3}+\frac{x+4}{x-6}+\frac{3 x-1}{x^{2}-3 x-18} $$

Step-by-Step Solution

Answer

\( \frac{3x^2 - 3x + 17}{(x+3)(x-6)} \)

Step 1: Identify the Common Denominator

First, we need to find the common denominator for the fractions. Notice that the third fraction's denominator can be factored: \(x^2-3x-18=(x+3)(x-6)\). Thus, the common denominator for all fractions is \( (x+3)(x-6)\).

Step 2: Rewrite Fractions with the Common Denominator

Convert each fraction to have the common denominator \( (x+3)(x-6) \): 1. \( \frac{2x-1}{x+3} = \frac{(2x-1)(x-6)}{(x+3)(x-6)} \)2. \( \frac{x+4}{x-6} = \frac{(x+4)(x+3)}{(x+3)(x-6)} \)3. \( \frac{3x-1}{(x+3)(x-6)} \) remains the same.

Step 3: Combine the Fractions

Now that the denominators are the same, we can add the numerators:\[ \frac{(2x-1)(x-6) + (x+4)(x+3) + (3x-1)}{(x+3)(x-6)} \] Expand each term in the numerator.

Step 4: Expand and Simplify the Numerators

Expand each term:\((2x-1)(x-6) = 2x^2 - 12x - x + 6 = 2x^2 - 13x + 6\)\((x+4)(x+3) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12\)Grouping with \(3x-1\), we have: \[ 2x^2 - 13x + 6 + x^2 + 7x + 12 + 3x - 1 \]

Step 5: Combine and Simplify

Combine like terms in the numerator: \( (2x^2 + x^2) + (-13x + 7x + 3x) + (6 + 12 - 1) \)This simplifies to:\( 3x^2 - 3x + 17 \)Thus, the expression in simplest form becomes: \[ \frac{3x^2 - 3x + 17}{(x+3)(x-6)} \]

Step 6: Final Expression

The simplified form of the given expression is: \[ \frac{3x^2 - 3x + 17}{(x+3)(x-6)} \] This is the simplest form, as the numerator cannot be factored further to cancel with terms in the denominator.

Key Concepts

Common Denominator

When adding rational expressions, similar to adding fractions, a common denominator is essential. It ensures the expressions have a uniform base to facilitate addition. The least common denominator (LCD) is typically the product of all unique factors from each fraction's denominator. In our exercise, the denominators initially were \(x+3\), \(x-6\), and a factored form of \(x^2-3x-18\) which is \((x+3)(x-6)\).
To find the LCD for these expressions:

• Factor any quadratic or complex denominators. Here, \(x^2-3x-18\) becomes \((x+3)(x-6)\).
• Identify all unique factors across the denominators: in this case, \((x+3)\) and \((x-6)\).
• The product of these unique factors \((x+3)(x-6)\) serves as the common denominator.

This step unifies the denominators allowing the addition of the numerators.

Factoring Quadratics

Factoring is a fundamental tool in algebra, especially when dealing with quadratic expressions. It involves breaking down a polynomial into simpler components, or factors, that when multiplied together give the original polynomial. In our case, we were faced with \(x^2-3x-18\).

Here's how we factor the quadratic:

• Identify the standard form: \(ax^2 + bx + c\), where here it is \(1x^2 - 3x - 18\).
• Look for two numbers that multiply to \(c = -18\) and add up to \(b = -3\). These numbers are \(-6\) and \(3\).
• Rewrite the expression as \((x-6)(x+3)\), the factors of the quadratic.

Factoring simplifies expressions and plays a crucial role in finding common denominators for rational expressions.

Simplifying Expressions

Simplifying an expression means reducing it to its simplest form, where no further factorization or cancellation is possible. After expanding and combining the terms in our exercise, we achieved a numerator of \(3x^2 - 3x + 17\) over \((x+3)(x-6)\).
This process involves:

• Expanding binomials and simplifying each component of the expression, like the numerators after distributing terms.
• Combining like terms by summing or subtracting coefficients with identical variables or constants.
• Checking if any further cancellation between the numerator and denominator is possible.

We ended with \(\frac{3x^2 - 3x + 17}{(x+3)(x-6)}\), in this case, which is fully simplified as the numerator cannot be factored further to reduce with the denominator.

Expanding Binomials

Expanding binomials involves applying the distributive property to remove parenthesis, a key part of simplifying quadratic expressions. This step is crucial for the addition of rational expressions as it aligns numerators with a common denominator.
Here's how expansion was approached in this exercise:

• The expression \((2x-1)(x-6)\) is expanded to: \(2x^2 - 12x - x + 6\), simplifying to \(2x^2 - 13x + 6\).
• The expression \((x+4)(x+3)\) is expanded to: \(x^2 + 3x + 4x + 12\), simplifying to \(x^2 + 7x + 12\).
• Addition of \(3x-1\) was made to the combined expansions.

This illustrates how expanding binomials makes further simplification possible and manageable, thereby setting the stage for combining like terms.