Fractions

We manipulate algebraic fractions the same way that we manipulate
fractions in arithmetic.

Simplifying Fractions

One fundamental rule for manipulation of fractions is: If we multiply
or divide both the numerator or denominator by the same quantity,
the value of the fraction will not change provided that this quantity
is nonzero, namely

$ \frac{A}{B}=\frac{AC}{BC},\qquad\frac{A}{B}=\frac{\dfrac{A}{C}}{\dfrac{B}{C}}\qquad(\text{if }C\neq0) $

For example, if we multiply both the numerator and the denominator
of $(x-1)/(x+2)$ by $(x-3)$, we obtain the equivalent fraction

$ \frac{x-1}{x+2}=\frac{(x-1)(x-3)}{(x+2)(x-3)} $

provided $x-3\neq0$; that is, provided $x\neq3$.

Conversely if we factor the numerator and denominator of a given fraction,
we can cancel common factors from the numerator and denominator to
simplify the fraction (again provided common factors are
not zero). For example,

$ \frac{4}{12}=\frac{\cancel{4}}{\cancel{4}\times3}=\frac{1}{3} $$ \frac{x^{2}-5x+6}{x^{2}-4x+3}=\frac{(x-2)\cancel{(x-3)}}{\cancel{(x-3)}(x-1)}=\frac{x-2}{x-1}\qquad(\text{if }x\neq3) $

Example

Simplify $\dfrac{x^{2}+x-12}{3-x}$

Solution

$\begin{aligned} \frac{x^{2}+x-12}{3-x} & =\frac{(x+4)(x-3)}{3-x} & =\frac{(x+4)\cancel{(x-3)}}{-\cancel{(x-3)}} & =-(x+4) & =-x-4. \end{aligned}$

Multiplying and Dividing Fractions

To multiply or divide fractions we use the following properties of
fractions:

|c|p{7cm}|} Property	Restriction	Description\tabularnewline $\dfrac{A}{B}\cdot\dfrac{C}{D}=\dfrac{A\cdot C}{B\cdot D}$	$B\neq0$, $D\neq0$	To multiply two fractions, multiply their numerators and multiply their denominators\tabularnewline $\dfrac{A}{B}\div\dfrac{C}{D}=\frac{\dfrac{A}{B}}{\dfrac{C}{D}}=\dfrac{A}{B}\cdot\dfrac{D}{C}$	$B\neq0,D\neq0$, $C\neq0$	To divide a fraction by another fraction, invert the divisor and then multiply the fractions.\tabularnewline

Example

Perform the following operations and simplify the results

(a) $\dfrac{x_{2}-2x-3}{x_{2}+6x+9}\cdot\dfrac{4x+12}{x+1}$

(b) $\dfrac{x^{2}-6x+8}{x^{2}-x-6}\div\dfrac{x^{2}-x-12}{4x-12}$

Solution

(a)

$\begin{aligned} \dfrac{x_{2}-2x-3}{x_{2}+6x+9}\cdot\dfrac{4x+12}{x+1} & =\frac{(x-3)\bcancel{(x+1)}}{(x+3)^{1\cancel{9}}}\times\frac{4\cancel{(x+7)}}{\bcancel{(x+1)}} & =\frac{4(x-3)}{x+3}, \end{aligned}$

provided $x+1\neq0$; that is provided $x\neq-1$.

(b)

$\begin{aligned} \dfrac{x_{2}-6x+8}{x_{2}-x-6}\div\dfrac{x_{2}-x-12}{4x-12} & =\dfrac{x^{2}-6x+8}{x^{2}-x-6}\cdot\dfrac{4x-12}{x^{2}-x-12} & =\frac{\bcancel{(x-4)}(x-2)}{\cancel{(x-3)}(x+2)}\cdot\frac{4\cancel{(x-3)}}{\bcancel{(x-4)}(x+3)} & =\frac{4(x-2)}{(x+2)(x+3)}, \end{aligned}$

provided $x\neq4$ and $x\neq3$.

Adding and Subtracting Fractions

If the denominators of the fractions are the same, we simply add or
subtract the numerators, namely

$ \frac{A}{B}\pm\frac{C}{B}=\frac{A\pm C}{B} $

If the denominators are different, we have to find a common denominator.
One way is to multiply the numerator and denominator of each fraction
by the denominator of the other one:

$\begin{aligned} \frac{A}{B}\pm\frac{C}{D} & =\frac{A}{B}\cdot\frac{D}{D}\pm\frac{C}{D}\cdot\frac{B}{B} & =\frac{A\cdot D\pm C\cdot B}{B\cdot D} \end{aligned}$

For example

$ \frac{5}{6}+\frac{4}{9}=\frac{5\cdot9+4\cdot6}{6\cdot9}=\frac{69}{54}=\frac{23}{18}. $

To make simplification easier, we often find the least common multiple
(LCM) of the denominators. The LCM of the denominators is called least
common denominator (LCD). For example, in the above example, the LCM of 6 and 9 is 18 (because
$6\times3=9\times2=18$), so

$ \frac{5}{6}+\frac{4}{9}=\frac{5\cdot3}{6\cdot3}+\frac{4\cdot2}{9\cdot2}=\frac{15+8}{18}=\frac{23}{18}. $

Least Common Multiple (LCM) and Least Common Denominator (LCD)

The least common multiple (LCM) of two or more polynomials
is the polynomial of lowest degree with smallest numerical coefficients
which is divisible by each of them (that is if we divide it by each
of the polynomials the remainder will be zero).

The least common multiple of the denominators of a set of fractions
is called the least common denominator (LCD).

How to find the LCM of two or more expressions:

1. Find the factors of each of the expression.
2. Select all of distinct factors and give to each the highest exponent
   with which it occurs in any of the expressions.
3. Find the product of all of the factors selected in step 2.

Example

Perform the operation and write in simplified form

$ \frac{4x}{x_{2}-4}+\frac{3x}{x_{2}-5x+6} $

Solution

The denominators are not the same, so we have to find LCM of them
(or LCD of the two terms)

$ \frac{4x}{x_{2}-4}+\frac{3x}{x^{2}-5x+6}=\frac{4x}{(x-2)(x+2)}+\frac{3x}{(x-3)(x-2)} $

so the LCD is $(x-2)(x+2)(x-3)$. We multiply the numerator and denominator
of the first fraction by $\dfrac{(x-2)(x+2)(x-3)}{(x-2)(x+2)}=(x-3)$
and the second one by $\dfrac{(x-2)(x+2)(x-3)}{(x-3)(x-2)}=(x+2)$
and then add the two fractions together:

$\begin{aligned} \frac{4x}{x^{2}-4}+\frac{3x}{x_{2}-5x+6} & =\frac{4x}{(x-2)(x+2)}+\frac{3x}{(x-3)(x-2)} & =\frac{4x(x-3)}{(x-2)(x+2)(x-3)}+\frac{3x(x+2)}{(x-3)(x-2)(x+2)} & =\frac{4x_{2}-12x+3x^{2}+6x}{(x-3)(x-2)(x+2)} & =\frac{7x_{2}-6x}{(x-3)(x-2)(x+2)} & =\frac{x(7x-6)}{(x-3)(x-2)(x+2)} \end{aligned}$

Compound Fractions

A compound fraction is a fraction that has one or more fractions in
the numerator or denominator or both. To simplify compound fractions:

Method 1:

Reduce the terms in the numerator and in the denominator into single
fractions. Then divide the two resulting fractions.

Method 2:

Find the least common multiple (LCM) of all denominators in the expression.
Then multiply the numerator and denominator by it.

Example

Simplify

$ \frac{\dfrac{1}{x_{2}}-4}{2-\dfrac{1}{x}} $

Solution

Method 1:

$\begin{aligned} \frac{\dfrac{1}{x_{2}}-4}{2-\dfrac{1}{x}} & =\frac{\dfrac{1-4x_{2}}{x_{2}}}{\dfrac{2x-1}{x}} & =\frac{1-4x_{2}}{x_{\cancel{2}}}\cdot\frac{\cancel{x}}{2x-1} & =\frac{1-(2x)_{2}}{x(2x-1)} & =\frac{\bcancel{(1-2x)}(1+2x)}{-x\bcancel{(x-2x)}} \quad (Using $A^{2-B^{2}=(A-B)(A+B)$) & =-\frac{1+2x}{x} \end{aligned}$

Method 2: The LCM of $x_{2}$ and $x$ is $x^{2}$

$\begin{aligned} \frac{\dfrac{1}{x_{2}}-4}{2-\dfrac{1}{x}} & =\frac{\dfrac{1}{x_{2}}-4}{2-\dfrac{1}{x}}\cdot\frac{x_{2}}{x_{2}} & =\frac{1-4x_{2}}{2x^{2}-x} & =\frac{(1-2x)(1+2x)}{x(2x-1)} & =-\frac{1+2x}{x} \end{aligned}$

Method 3: We note that

$\begin{aligned} \frac{1}{x_{2}}-4 & =\left(\frac{1}{x}\right)^{2}-2^{2} & =\left(\frac{1}{x}-2\right)\left(\frac{1}{x}+2\right) \quad (Using $A_{2}-B_{2}=(A-B)(A+B)$) \end{aligned}$

Thus

$\begin{aligned} \frac{\dfrac{1}{x^{2}}-4}{2-\dfrac{1}{x}} & =\frac{\left(\frac{1}{x}-2\right)\left(\frac{1}{x}+2\right)}{-\left(\frac{1}{x}-2\right)} & =-\left(\frac{1}{x}+2\right) & =-\frac{1+2x}{x}. \end{aligned}$

Example

Simplify

$ \frac{(1+x)^{1/2}-x(1+x)^{-1/2}}{x+1} $

Solution

We may rewrite the given fraction as

$ \frac{\sqrt{1+x}-\dfrac{x}{\sqrt{1+x}}}{x+1} $

Method 1:

$\begin{aligned} \frac{\sqrt{1+x}-\dfrac{x}{\sqrt{1+x}}}{x+1} & =\frac{\dfrac{(\sqrt{1+x})^2-x}{\sqrt{1+x}}}{x+1} & =\frac{\dfrac{1+x-x}{\sqrt{1+x}}}{x+1} & =\frac{1}{(x-1)\sqrt{1+x}} \end{aligned}$

which can also be written as

$\frac{1}{\sqrt{(1+x)^}3}}\qquad\text{or}\qquad\frac{1}{(1+x)^{3/2}}$

Method 2: Multiply both numerator and denominator by $\sqrt{1+x}$

$\begin{aligned} \frac{\sqrt{1+x}-\dfrac{x}{\sqrt{1+x}}}{x+1} & =\frac{\sqrt{1+x}-\dfrac{x}{\sqrt{1+x}}}{x+1}\cdot\frac{\sqrt{1+x}}{\sqrt{1+x}} & =\frac{(1+x)\cancel{(x-1)}}{(1+x)\sqrt{1+x}} & =\frac{1}{(1+x)\sqrt{1+x}} & =\frac{1}{(1+x)^}3/2}. \end{aligned}$