Problem 41 - Introductory Algebra for College Students

Question

Simplify complex rational expression. \(\frac{\frac{6}{x^{2}+2 x-15}-\frac{1}{x-3}}{\frac{1}{x+5}+1}\)

Step-by-Step Solution

Answer

The simplified form of the complex rational expression is \(\frac{1-x}{x^2+5x+5}\), where \(x ≠ -5, 3\).

Step 1: Simplify the Expressions

First, factor the quadratic equation in the denominator of the first fraction in the numerator, \(x^{2}+2 x-15\) can be factored into \((x+5)(x-3)\). The expression then becomes \(\frac{\frac{6}{(x+5)(x-3)}-\frac{1}{x-3}}{\frac{1}{x+5}+1}\)

Step 2: Find the LCD

In order to simplify the expression, find the least common denominator (LCD). The LCD for \((x+5)(x-3)\) and \(x-3\) is \((x+5)(x-3)\). Multiply both terms in the numerator by this LCD to get rid of fractions. The expression then becomes \(\frac{6-(x+5)}{(x+5)+ (x+5)(x-3)}\) which simplifies to \(\frac{1-x}{x^2+5x+5}\)

Step 3: Check for Undefined Values

Finally, be aware that the given expression is undefined for \(x = -5\) and \(x = 3\). Always exclude the values that make the expression undefined.

Key Concepts

Factoring Quadratics

Factoring quadratics is the process of breaking down a quadratic expression into simpler expressions, usually in the form of binomials that multiply together to give the original quadratic. Understanding this concept is essential, especially when dealing with complex rational expressions.
Quadratics are typically in the form of \( ax^2 + bx + c \). For our example, the expression is \( x^2 + 2x - 15 \).

To factor this expression, we are looking for two numbers that multiply to \(-15\) (the constant term) and add to \(2\) (the linear coefficient). Those numbers are \(5\) and \(-3\). This makes it possible to express the quadratic as:

• \( x^2 + 2x - 15 = (x + 5)(x - 3) \)

Factoring quadratics not only simplifies the expression but is also crucial in identifying potential restrictions, like undefined values, which are pivotal in evaluating rational expressions safely.

Least Common Denominator

In solving complex rational expressions, finding the least common denominator (LCD) is crucial. It allows us to combine fractions by creating a common base which simplifies addition or subtraction across different denominators.
In the example problem, the denominators are \( (x+5)(x-3) \) and \( x-3 \). The LCD must include every unique factor at least as many times as it appears in the original denominators.

Here, the LCD is \((x+5)(x-3)\). By multiplying both terms in the numerator by this LCD, we eliminate the fractions, making subsequent operations straightforward.
This step is essential because it simplifies the expression to a form where you can easily perform addition and subtraction, ultimately leading towards a more manageable expression.

Undefined Values

Undefined values in expressions occur when the denominator is zero, as division by zero is mathematically undefined. Identifying these values ensures that any solution to the rational expression respects mathematical rules.
In our expression, taking the factors from the denominator \((x+5)(x-3)\), setting them individually equal to zero helps find the points where the expression becomes undefined. This gives:

• For \(x+5 = 0\): \(x = -5\)
• For \(x-3 = 0\): \(x = 3\)

Therefore, the expression does not exist or is undefined at \(x = -5\) and \(x = 3\). Being aware of these values is essential since they restrict the domain of the function and ought to be excluded from any final solution or interpretation of the expression.