Problem 42 - Introductory Algebra for College Students

Question

Simplify complex rational expression. \(\frac{\frac{1}{x-2}-\frac{6}{x^{2}+3 x-10}}{1+\frac{1}{x-2}}\)

Step-by-Step Solution

Answer

The simplified form of the complex fraction is \(\frac{x-2}{x+5}\).

Step 1: Simplify the denominator of the big fraction.

Find a common denominator for \(1\) and \(\frac{1}{x-2}\). This is \(x-2\). This leads to \(1 + \frac{1}{x-2} = \frac{x-2}{x-2} + \frac{1}{x-2} = \frac{x-2+1}{x-2} = \frac{x-1}{x-2}\).

Step 2: Simplify the numerator of the big fraction.

Find a common denominator for \(\frac{1}{x-2}\) and \(\frac{6}{x^{2}+3 x-10}\). First decompose \(x^{2}+3 x-10\) into \((x-2)(x+5)\). Therefore, \(\frac{1}{x-2}\) and \(\frac{6}{(x-2)(x+5)}\) have the common denominator \((x-2)(x+5)\). When these are added, we get \(\frac{x+5-6}{(x-2)(x+5)} = \frac{x-1}{(x-2)(x+5)}\).

Step 3: Simplify the big fraction.

This gives \(\frac{\frac{x-1}{(x-2)(x+5)}}{\frac{x-1}{x-2}}\). Dividing by a fraction is equal to multiplying by its reciprocal. Therefore this equals \(\frac{x-1}{(x-2)(x+5)}*\frac{x-2}{x-1} = \frac{x-2}{x+5}\), as the \(x-1\) cancels out in the numerator and the denominator.

Key Concepts

Simplifying Fractions

Simplifying fractions is like tidying up a messy room; you're trying to make everything neat and straightforward. When faced with fractions, especially complex ones, our goal is to simplify them to a form that's easier to understand and work with.

• Start by identifying the common elements in the numerator and the denominator.
• These can be numbers or algebraic expressions that we can cancel out.

This process often involves finding a common denominator, making sure all terms include it, and then canceling out similar terms. In our example, simplifying means we have to deal with the complex fraction, such as \(rac{x-1}{(x-2)(x+5)}\) over another simpler fraction \(rac{x-1}{x-2}\).
When we simplify, we multiply by the reciprocal of the denominator, which neatens the fraction up. Remember, the aim is to rid the fraction of any unnecessary complexity.

Common Denominator

The concept of a common denominator is fundamental when dealing with multiple fractions. Having a common denominator allows us to add or subtract fractions easily.
For instance, in the expression \(rac{1}{x-2}+rac{1}{x-2}\), the common denominator is simply \(x-2\).
Whenever you work with more complex fractions involving variables, like the exercise here, finding the common denominator can be more intricate. We are dealing with \(\frac{1}{x-2}\) and \(\frac{6}{(x-2)(x+5)}\) where the common denominator becomes \((x-2)(x+5)\).

• This step is crucial as it paves the way for combining the fractions.
• It ensures each fraction is represented consistently, which is essential for further simplification or addition.

Much like finding a common language to communicate, a common denominator brings fractions onto a level playing field, making it easier to work with them.

Polynomial Factorization

Polynomial factorization is a skill that turns a complex polynomial into simpler, more manageable pieces. Think of it as breaking down a large building into its smaller, basic blocks.

• The purpose of factorization is to simplify polynomials to the point where they can more easily be worked with in equations.
• You identify factors common to different terms and decompose the polynomial accordingly.

In the exercise, we are asked to factor the polynomial in the denominator, \(x^2 + 3x - 10\).
By breaking it down, we get \((x-2)(x+5)\).
This step is crucial, as it enables you to simplify fractions and expressions by revealing shared factors that can be canceled out. Much like unraveling a knot, factorization shows us the simplest way to understand and manipulate an algebraic expression.