Algebra

Rational Expressions

Part 1: Addition and Subtraction

Consider simplifying the following rational expression: $$ \frac{1}{x+1}-\frac{2}{(x+1)^2}+\frac{3}{x^2-1} $$ Let’s begin by recalling our approach with expressions that only involve numbers. We know that in order to add or subtract two fractions, we must first have a common denominator. Below is a common property to help achieve that goal: $$\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$$ Although this property guarantees that we get a common denominator, we may or may not have found the Least Common Denominator (LCD). So let’s take a moment to explore how to find it since LCDs have the potential to drastically simplify our computations later on.
Consider the following difference: $\frac{11}{12}-\frac{7}{10}$
Using the above property, we have: $$\begin{align}\frac{11}{12} - \frac{7}{10} &= \frac{11(10)-7(12)}{12(10)} \\ &= \frac{110-84}{120} \\ &= \frac{26}{120} \\ &= \frac{13}{60} \end{align}$$ Notice that $120$ is twice as large as the reduced denominator of $60$! If you found it obvious that the LCD was $60$, note that the LCD is not always so easy to identify. So, below is a systematic approach to finding it.

Step 1. Factor each denominator (i.e. express each denominator as a product of its prime factors).
 For example:
$$ 12 = 2^2\cdot 3\qquad \qquad \qquad 10 = 2 \cdot 5 $$
Step 2: Construct the LCD by multiplying together the highest power of each prime that appears among the denominators. $$\textrm{LCD } = 2^2\cdot 3 \cdot 5 = 60$$ Step 3: Multiply each fraction’s numerator and denominator by the prime factors missing in its denominator and simplify the expression. $$\frac{11(5)-7(2\cdot 3)}{60} = \frac{55-42}{60}=\frac{13}{60}$$ Now, because variables are just numbers in disguise, let’s use this numerical example to direct our approach for the rational expression first proposed: $$ \frac{1}{x+1}-\frac{2}{(x+1)^2}+\frac{3}{x^2-1} $$ Step 1: Factor the denominators

Denominator 1: $ \quad x+1$	(already fully factored)
Denominator 2: $\quad (x+1)^2$	(already fully factored)
Denominator 3: $ \quad x+1 = (x+1)(x-1)$	(Difference of Squares; requires factoring)

Step 2:
LCD: $(x+1)^2(x-1)$

Step 3:
$$\begin{align*} \frac{1\cdot (x+1)(x-1) - 2\cdot (x-1) + 3\cdot (x+1)}{(x+1)^2(x-1)} &= \frac{x^2-1-2x+2+3x+3}{(x+1)^2(x-1)} \\ &= \frac{x^2+x+4}{(x+1)^2(x-1)} \end{align*}$$

Part 2: Multiplication and Division

Recall that to multiply two fractions, we simply multiply the numerators together and write the product on top. Then multiply the denominators together and write the product on the bottom. $$\textrm{i.e. }\frac{a}{b}\frac{c}{d}= \frac{ac}{bd} $$ Division, on the other hand, is performed by multiplying by the reciprocal (the fraction where the top and bottom are flipped). $$\textrm{i.e. }\frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}= \frac{a}{b} \div \frac{c}{d} = \frac{a}{b}\cdot\frac{d}{c} = \frac{ad}{bc} $$ Consider the following example: $$\frac{x+3}{x^2+7x+10}\left(\frac{x+3}{x-5}\right)\div \frac{x+3}{x^2-25} $$ TIP: Begin by factoring the expressions in the numerators and denominators because it will make it easier to simplify the expression at the end! $$\begin{align*} \frac{x+3}{x^2+7x+10}\left(\frac{x+3}{x-5}\right)\div \frac{x+3}{x^2-25} &= \frac{x+3}{(x+2)(x+5)}\left(\frac{x+3}{x-5}\right)\div \frac{x+3}{(x-5)(x-5)} \\ &= \frac{(x+3)(x+3)}{(x+2)(x+5)(x-5)}\cdot \frac{(x-5)(x+5)}{(x+3)} \\ &= \frac{(x+3)(x+3)(x+5)(x-5)}{(x+3)(x+2)(x+5)(x-5)} \\ &= \frac{\cancel{(x+3)}(x+3)\cancel{(x+5)}\cancel{(x-5)}}{\cancel{(x+3)}(x+2)\cancel{(x+5)}\cancel{(x-5)}} \\ &= \frac{(x+3)}{(x+2)} \end{align*}$$ It is important to note that the only reason we were able to cancel those brackets is because $\frac{(x+3)}{(x+3)} = \frac{(x+5)}{(x+5)} = \frac{(x-5)}{(x-5)} = 1$, and multiplying by 1 does not change the value of our final solution!