Manipulating Algebraic Expressions and Fractions with Excluded Values

1. What Makes a Rational Expression Undefined?

A rational expression is simply a fraction where the numerator and denominator are polynomials. For example, $\frac{x+1}{x-2}$ and $\frac{x^2+7x+12}{x^2-9}$ are rational expressions.

The single most important rule when dealing with fractions of any kind is that the denominator can never be zero. Division by zero is undefined. Therefore, before we perform any operation, our first thought should always be: "What values of the variable are not allowed?" These are the excluded values. Finding them is equivalent to determining the function's domain, a concept that is central to many areas of mathematics.

Let's start by looking at the formal process for identifying these values.

As the resource shows, finding excluded values is a matter of setting the denominator equal to zero and solving for the variable. For a denominator like $x^2-9$, you would solve $x^2-9=0$, which gives $x=3$ and $x=-3$ as the excluded values.

2. Simplifying Rational Expressions

Just as we can simplify the numerical fraction $\frac{6}{9}$ to $\frac{2}{3}$, we can simplify rational expressions. The process is identical: find common factors in the numerator and denominator, and then cancel them out. This requires you to be proficient in factoring polynomials—a skill we will continually reinforce.

A common and costly error is to cancel terms instead of factors. For example, in the expression $\frac{x+3}{x}$, you cannot cancel the $x$'s. The $x$ in the numerator is a term being added to 3, not a factor multiplying the entire numerator. However, in $\frac{x(x+3)}{x}$, you can cancel the $x$'s, because here it is a factor.

The following reading details the correct procedure.

A Special Case: Opposite Factors

Sometimes, you'll encounter factors that are opposites, like $(x-5)$ and $(5-x)$. Recognizing this pattern can simplify your work. Since $(5-x)=-(x-5)$, their ratio is always -1.

This is a neat trick that frequently appears in IB questions. Let's see how it's formally applied.

3. Adding and Subtracting Rational Expressions

To add or subtract fractions, you must have a common denominator. Once you do, you simply combine the numerators. When subtracting, be especially careful to distribute the negative sign to every term in the second numerator.

If the denominators are not the same, you must first find the Least Common Denominator (LCD). The LCD is the smallest polynomial that is a multiple of all the individual denominators. The most effective way to find it is to factor each denominator completely and then build a new denominator that contains all of these factors, with each factor raised to its highest power.

4. Complex Fractions

A complex fraction is a fraction that contains other fractions in its numerator, denominator, or both. They may look intimidating, but they are just a combination of the skills we've already covered.

For example:

The strategy is to simplify the numerator into a single fraction and the denominator into a single fraction. Once you have a single fraction divided by another, you can multiply the top fraction by the reciprocal of the bottom one.

The following video provides an excellent walkthrough of a challenging example. Pay close attention to how the presenter handles the excluded values at the end—this is a crucial point for IB exams.

The key insight from the video is that an excluded value is any value that makes any denominator zero at any stage of the original, unsimplified expression.

IB-Style Practice

Let's apply these skills to a typical IB-style problem structure.

Problem:

Let $f(x)=\frac{2x-1}{x^2-x-2}$ and $g(x)=\frac{1}{x-2}$.

(a) State the values of $x$ for which $f(x)$ is undefined.
(b) Show that $f(x)-g(x)=\frac{x}{x^2-x-2}$.
(c) Find the domain of the function $h(x)=f(x)-g(x)$.

Your Turn:

Take some time to work through this problem. Think about how you would present your solution clearly, step-by-step, especially for part (b) where you must "show that".

Solution:

(a) To find where $f(x)$ is undefined, we set its denominator to zero.
$x^2-x-2=0$
$(x-2)(x+1)=0$
So, $x=2$ or $x=-1$.
$f(x)$ is undefined for $x=2$ and $x=-1$.

(b) To show the result, we start from the left-hand side and manipulate it until it looks like the right-hand side.

The LCD is $(x-2)(x+1)$. We need to rewrite the second fraction with this denominator.

Wait, this does not match the target expression $\frac{x}{x^2-x-2}$. This is a good moment to pause and check our work. It seems there was a typo in the original problem statement. Let's assume the target should have been $\frac{1}{x+1}$ based on our correct algebraic work. The key is that the process is sound. Self-correction is a vital skill.

Let's re-examine the subtraction step, a common place for errors.
$(2x-1)-(x+1)=2x-1-x-1=x-2$. The algebra is correct.

Tutor's Note: It seems I made an error when writing the problem. My apologies! Let's correct the problem so it works. Let's change $g(x)$ to $g(x)=\frac{1}{x+1}$. Now let's try part (b) again.

Corrected Problem:
Let $f(x)=\frac{2x-1}{x^2-x-2}$ and $g(x)=\frac{1}{x+1}$.
Show that $f(x)-g(x)=\frac{1}{x-2}$.

Solution 2.0:

This works perfectly. The "show that" is complete.

(c) To find the domain of $h(x)$, we must look at the original functions, $f(x)$ and $g(x)$. The domain of $h(x)$ includes only values for which both $f(x)$ and $g(x)$ are defined.
From part (a), $f(x)$ is undefined for $x=2,x=-1$.
For our corrected $g(x)=\frac{1}{x+1}$, it is undefined when $x=-1$.
Combining these, the excluded values are $x=2$ and $x=-1$.
Therefore, the domain of $h(x)$ is all real numbers except 2 and -1. In set notation, this is $\{x\in \mathbb{R}\mid x\ne -1,x\ne 2\}$.

Conclusion

In this lesson, we have established the foundational skills for working with rational expressions. You now have a systematic process for simplifying, adding, and subtracting them.

Here are the key takeaways:

• Excluded Values are Paramount: Always start by finding the values that make any denominator in the original expression zero.
• Factor, then Cancel: Simplification only works on common factors, not terms. Factor everything first.
• LCD is Key for Addition/Subtraction: Find the LCD by factoring all denominators, then build equivalent fractions.
• The Domain is Inherited: The domain of a simplified or combined expression is determined by the excluded values of the original expressions.

Having a strong command of these algebraic manipulations is not just about getting algebra questions right; it's about freeing up your mental energy to tackle the more complex concepts in calculus, where these skills will be used constantly.

In our next lesson, we will continue to strengthen our algebraic fluency by tackling another core topic: solving equations that involve indices and logarithms.