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Section 6.1 Fractions and Variables

Subsection 6.1.1 Fractions with Variables

Figure 6.1.1. Factions with Variables; Factoring

One we start working with variables, functions and related mathematical expressions, all of these things can show up in fractions.

\begin{equation*} \frac{\sqrt{t^2+5} + t}{t^2 \cos t} \end{equation*}

As we will see, the rules of fraction arithmetic remain the same, though some of the processes become more cumbersome and complicated with variable expressions.

With variables and mathematical expressions, we also have to worry about potential division by zero. In the fraction

\begin{equation*} \frac{6y-5}{y+3}\text{,} \end{equation*}

all values of \(y\) except for \(y = -3\) are acceptable, but \(y = -3\) would lead to division by zero. It is very valuable to always keep in mind potential division by zero and take note of the values of variables that must be avoided.

For fractions with variables, we can still multiply numerator and denonminator by the same thing and preserve the fraction.

\begin{equation*} \frac{\sin z}{z} = \frac{(\sin z) (z^2)}{(z)(z^2)} = \frac{z^2 \sin z}{z^3} \end{equation*}

However, this is only true is we haven't multiplied by zero. We need to keep in our mind that this expression doesn't work for \(z = 0\text{.}\) If we had multiplied numerator and denominator by \((z - 2)\) instead, then we would have to keep in mind that are new expression is not valid for \(z = 2\text{.}\)

These new fractions don't necessary have any special terminology, except for one case. When the numerator and denominator are both polynomials (as discussed in Chapter 13), the fraction is called a rational function. For example

\begin{equation*} \frac{k^2 + 3k - 4}{7k^3 - 4} \end{equation*}

is a rational function (in the variable \(k\)) and

\begin{equation*} \frac{a^7 + 4a^3 - 9a}{a^5 - 35} \end{equation*}

is a rational function (in the variable (\(a\)). However,

\begin{equation*} \frac{\sqrt{x^2+1}}{x-1} \end{equation*}

is not a rational function, since the square root is not a polynomial.

Most of the examples in this section, and most of the problems in the activity, will be using rational functions. If might be useful to do this section along side Chapter 13. I'll use some of the techniques review in that chapter, such as the multiplication of binomial and the simplification of some polynomial expressions.

Subsection 6.1.2 Fractions and Factoring

For fraction with variables, we don't really talk about lowest terms anymore (that terminology is restricted to numbers, where there is a clearer idea of a simplest form). However, we can still factor terms out of the numerator and denominator of a fraction to simplify the expression.

In the first part, we can still factor numbers out the numerator and denmoninator.

\begin{equation*} \frac{8x^2 + 4x}{2 + 10\sqrt{x}} \end{equation*}

In this expression, we can factor \(2\) out of both numerator and denominator.

\begin{equation*} \frac{2(4x^2 + 2x)}{2(1 + 5\sqrt{x})} = \frac{4x^2 + 2x}{1 + 5 \sqrt{x} \end{equation*}

Then we can divide numerator and denominator by \(2\text{.}\) We often don't explicitly show this division, but just cross off the \(2\) from each piece. We often refer to this as cancelling off.

More substantially, we can also factor our and cancel off expressions with variables.

\begin{equation*} \frac{t^2 + 8t}{5t^2 - 4t} \end{equation*}

In this rational function, we can factor out \(t\) from both numerator and denominator.

\begin{equation*} \frac{t(t + 8)}{t(5t - 4)} \end{equation*}

Then we can cancel of this \(t\text{.}\) Implicitly, when we do this cancelation, we are dividing numerator and denominator by \(t\text{.}\)

\begin{equation*} . \frac{t(t + 8)}{t(5t - 4)} = \frac{t+8}{5t-4} \end{equation*}

Here there is one caution and subtlety: the previous equality is only true if \(t \neq 0\text{.}\) Much like multiplying numerator and denominator by a term, when we cancel of a term, we need to make sure the term was non-zero. If \(t=0\text{,}\) the left side of the above equation is undefined due to division by zero, but the right side nicely evaluates to \((-2)\text{.}\) That violates the equality. Therefore, we have to take note of the addition posibilities that may arise if the cancelled term was actually zero.

Subsection 6.1.3 More Fraction Arithmetic

Figure 6.1.2. Fraction Arithmetic with Variables

Section 5.1 reviewed the rules of fraction arithmetic. The examples in the section where all numbers. However, the rules of fraction arithmetic are the same for expressions involving variables as well.

Multiplication of fraction is still multiplication of numerators and mulitplication of denominators.

\begin{equation*} \left(\frac{x^2}{x^2+1}\right) \left(\frac{2^x}{\cos x} \right) = \frac{(x^2)(2^x)}{(x^2+1)\cos x} \end{equation*}

Likewise, division of fractions is still multiplication by reciprocals.

\begin{equation*} \frac{\frac{y^2+1}{y}}{\frac{y^3-6}{y}} = \left( \frac{y^2+1}{y} \right) \left( \frac{y}{y^3-6} \right) = \frac{(y^2+1)(y)}{(y)(y^3-6)} \end{equation*}

There is a factor of \(y\) in numerator and denominator, so as long as \(y \neq 0\text{,}\) we can factor it out and cancel it off.

\begin{equation*} \frac{(y)(y^2+1)}{(y)(y^3-6)} = \frac{y^2+1}{y^3-6} \end{equation*}

This is the result of the division when \(y \neq 0\text{.}\) (When \(y = 0\text{,}\) the original expression was not defined due to division by zero, so this division holds for all possible \(y\) after all).

Addition and subtraction of fractions with variables still required common denominator. However, the common denominator is now (usually) an expression with variables.

\begin{equation*} \frac{r + 1}{r^2} + \frac{r - 4}{r - 1} \end{equation*}

As with numbers, we can also take the common denominator to be the product of the two denominator. We multiply the numerator and denominator of the first fraction by \((r-1)\text{;}\) likewise, we multiply the numerator and denominator of the second fraction by \(r^2\text{.}\)

\begin{equation*} \frac{(r + 1)(r - 1)}{r^2(r - 1)} + \frac{(r - 4)(r^2)}{(r - 1)(r^2)} \end{equation*}

That produces a common demoniator of \(r^2(r-1)\text{.}\) We can expand the multiplications in the numerator to see what we get.

\begin{equation*} \frac{r^2 - 1}{r^2(r - 1)} + \frac{r^3 - 4r^2}{r^2(r - 1)} = \frac{r^2 - 1 + r^3 - 4r^2}{r^2(r - 1)} = \frac{r^3 - 3r^2 - 1}{r^2(r - 1)} \end{equation*}

Since we multiplied by both\(r\) and \((r - 1)\text{,}\) this calcluation is only valid when \(r \neq 0\) and \(r \neq 1\text{.}\)

As with common denominator with numbers, sometimes there is an easier common denominator than simply the product of the two denominators. An example will illustrate this.

\begin{equation*} \frac{6}{v^2(v+1)} - \frac{8}{v(v+1)(v+2)} \end{equation*}

The two fractions here have reasonable complicated denominators. I'd like to avoid multiplying by the whole denominator if I can. Looking at the two denominators, I see some common factors: \(v\) and \((v+1)\text{.}\) The first denominator is missing the \((v+2)\) in the second, so if I multiply by \((v+2)\) I get \(v^2(v+1)(v+2)\text{.}\) The second denominator is missing the square of the variables, so if I multiply by \(v\text{,}\) I get \(v^2(v+1)(v+2)\text{.}\) In this way I can produce a common denominator without introducing as much complexity to the situation. Let's proceed with the subtraction, multiplying the numerator and denominator of the first fraction by \((v+2)\text{,}\) and the second fraction by \(v\text{.}\)

\begin{equation*} \frac{6(v+2)}{v^2(v+1)(v+2)} - \frac{8v}{v^2(v+1)(v+2)} \end{equation*}

This produces a common denominator, so I can combine the fractions and do the subtraction.

\begin{equation*} \frac{6(v+2) - 8v}{v^2(v+1)(v+2)} \end{equation*}

I can expand the \(6(v+2)\text{.}\)

\begin{equation*} \frac{6v + 12 - 8v}{v^2(v+1)(v+2)} = \frac{-2v + 12}{v^2(v+1)(v+2} \end{equation*}

I don't see any more simplification to do here, so this is the result of our subtraction. We are avoiding \(v = 0\text{,}\) \(v = -1\) and \(v = -2\) in this calculation to avoid division by zero.