# Algebra

## Worked Examples

## 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!