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,
\[\frac{4}{12}=\frac{\cancel{4}}{\cancel{4}\times3}=\frac{1}{3}\]
and
\[\frac{x^{2}-5x+6}{x^{2}-4x+3}=\frac{(x-2)\cancel{(x-3)}}{\cancel{(x-3)}(x-1)}=\frac{x-2}{x-1}\qquad(\text{if }x\neq3)\]
Example 1
Simplify $\dfrac{x^{2}+x-12}{3-x}$
Solution
\begin{align*}
\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{align*}
Multiplying and Dividing Fractions
To multiply or divide fractions we use the following properties of fractions:
| Property |
Restriction |
Description |
| $\dfrac{A}{B}\cdot\dfrac{C}{D}=\dfrac{A\cdot C}{B\cdot D}$ |
$B\neq0$, $D\neq 0$ |
To multiply two fractions, multiply their numerators and multiply their denominators |
| $\dfrac{A}{B}\div\dfrac{C}{D}=\frac{\dfrac{A}{B}}{\dfrac{C}{D}}=\dfrac{A}{B}\cdot\dfrac{D}{C}$ |
$B\neq0,D\neq0$, $C\neq 0$ |
To divide a fraction by another fraction, invert the divisor and then multiply the fractions. |
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{align*}
\dfrac{x^{2}-2x-3}{x^{2}+6x+9}\cdot\dfrac{4x+12}{x+1} & =\frac{(x-3)\bcancel{(x+1)}}{(x+3)^{\cancel{2}}}\times\frac{4\cancel{(x+3)}}{\bcancel{x+1}}\\
& =\frac{4(x-3)}{x+3},
\end{align*} provided $x+1\neq0$; that is, provided $x\neq-1$.
(b)
\begin{align*}
\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{align*} 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{align*}
\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{align*}
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 real numbers, we know how to find LCM.
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:
- Find the factors of each of the expression.
- Select all of distinct factors and give to each the highest exponent with which it occurs in any of the expressions.
- 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 $\frac{(x-2)(x+2)(x-3)}{(x-2)(x+2)}=(x-3)$ and the second one by $\frac{(x-2)(x+2)(x-3)}{(x-3)(x-2)}=(x+2)$ and then add the two fractions together:
\begin{align*}
\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{align*}
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{align*}
\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{(1-2x)}}\\
& =-\frac{1+2x}{x}
\end{align*}
[ We used the identity $A^2-B^2=(A-B)(A+B)$, we wrote $1-(2x)^2=(1-2x)(1+2x)$]
Method 2: The LCM of $x^{2}$ and $x$ is $x^{2}$
\begin{align*}
\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{align*}
Method 3: This method works only for this specific problem. We note that
\begin{align*}
\frac{1}{x^{2}}-4 =\left(\frac{1}{x}\right)^{2}-2^{2}
\end{align*}
Now let $A=\frac{1}{x}$ and $B=2$. Using the identity $A^2-B^2=(A-B)(A+B)$, we can write
\begin{align*}
\left(\frac{1}{x}\right)^{2}-2^{2} =\left(\frac{1}{x}-2\right)\left(\frac{1}{x}+2\right)
\end{align*}
Thus
\begin{align*}
\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{align*}
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{align*}
\frac{\sqrt{1+x}-\dfrac{x}{\sqrt{1+x}}}{x+1} & =\frac{\dfrac{(\sqrt{1+x})^{2}-x}{\sqrt{1+x}}}{x+1}\\[12pt]
& =\frac{\dfrac{1+x-x}{\sqrt{1+x}}}{x+1}\\[12pt]
& =\frac{1}{(x+1)\sqrt{1+x}}
\end{align*}
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 the numerator and denominator by $\sqrt{1+x}$
\begin{align*}
\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)-\dfrac{x\cancel{\sqrt{1+x}}}{\cancel{\sqrt{1+x}}}}{(1+x)\sqrt{1+x}}\\
& =\frac{1}{(1+x)\sqrt{1+x}}\\
& =\frac{1}{(1+x)^{3/2}}.
\end{align*}
