Splitting and writing with a common denominator

Calculating with letters: Fractions with letters

Splitting and writing with a common denominator

Rational expressions

We have already discussed the rules for calculating with ordinary fractions. These rules remain valid when letters occur in fractions because these fractions with letters become ordinary fractions as soon as you substitute values for the variables. We speak of rational expressions. The only thing you have to be cautious for is that the denominator may not be zero.

Examples

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

$\frac{a+b}{a^2+b^2}$

Equivalence of expressions The numerator and denominator in a rational expression can be simultaneously divided by a nonzero number in order to simplify the mathematical expression. For example

$\frac{2x^2+2x}{2x}=\frac{x^2+x}{x}$ Whatever value you fill in for $x$ , the result is for both expressions the same. We call them equivalent expressions.

In the right-hand side of the expression in the example you cannot fill in $x=0$ , but for any other substitution you can divide both the numerator and denominator by $x$ and get

$\frac{x^2+x}{x}=x+1\text.$ The right-hand side of this expression is obviously easier, in strict mathematical sense $\dfrac{x^2+x}{x}$ and $x+1$ are different mathematical expression because they are subject to different conditions.

Henceforth we will treat such conditions loosely and do not often mention them explicitly. Silently we assume that the numeric values of the variables, when selected, stay outside the 'forbidden' region.

The simplification

$\frac{x}{x^2+x}=\frac{1}{x+1}$ is only valid if $x\neq 0$ and $x\neq -1$ because otherwise you divide the expression on the left-hand side by zero. But the condition $x\neq 0$ is not needed anymore for the expression on the right-hand side. In a reduction of $\dfrac{x}{x^2+x}$ to $\dfrac{1}{x+1}$ nothing goes wrong because the simplified expression is valid for all numbers that meet the conditions of the original expression; no new conditions were secretly added. However, when you read the equality from right to left and multiply the numerator and denominator by $x$ , there is a problem because a new condition must be added, namely $x\neq 0$ .

It depends on the problem or situation when you explicitly take the condition into account and write down for which values of variables all expressions are well-defined. If $x$ is a symbol for a quantity that can only take positive values, then the formula manipulation

$\frac{x}{x^2+x}=\frac{1}{x+1}$ can be done without wasting many words on the underlying conditions for simplification. Then dividing the numerator and denominator by $x$ does not cause any problems.

Addition and difference of algebraic fractions
Addition and subtraction of fractions with letters is not different from these operation for ordinary fraction: If the fractions have the same denominator in a sum or difference, then the denominator of the outcome is the same and the only thing you need to do is adding or subtracting the numerators. Otherwise, you must first write the fractions with a common denominator. Two examples:

$\begin{aligned}\frac{1}{a}+\frac{1}{b}&=\frac{b}{ab}+\frac{a}{ab}\\[0.25cm] &=\frac{a+b}{ab}\end{aligned}$

$\begin{aligned}\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}&=\frac{x}{xyz}+\frac{y}{xyz}+\frac{z}{xyz}\\[0.25cm] &=\frac{x+y+z}{xyz}\end{aligned}$

Splitting algebraic fractions
What also is often done in practice is splitting of algebraic fractions. Two examples:

$\begin{aligned}\frac{x^2+1}{x}&=\frac{x^2}{x}+\frac{1}{x}\\[0.25cm]&=x+\frac{1}{x}\end{aligned}$

$\begin{aligned}\frac{1}{x(x-1)}&=\frac{x-(x-1)}{x(x-1)}\\[0.2cm] &=\frac{x}{x(x-1)}-\frac{x-1}{x(x-1)}\\[0.2cm] &=\frac{1}{x-1}-\frac{1}{x}\end{aligned}$

In the example $\dfrac{x^2+1}{x}=x+\dfrac{1}{x}$ you notice from the splitting of the algebraic fraction that when you substitute a large number in $x$ the outcome hardly differs from the number itself. The splitting of the algebraic fraction reveals here the structure of the algebraic expression.