Finding a common denominator

Algebra: Rational Expressions

Finding a common denominator

When calculating with numbers we already encountered rules for calculating with fractions. These rules remain valid when variables appear in the fractions, just because they turn into normal fractions when we replace the variables by numbers. In the theory Simplifying fractions we saw that it might nolonger be possible to tell from the simplified fraction where the denominator of the original fraction is equal to zero.

Addition and subtraction of fractions

Addition and subtraction of fractions with variables is just like regular addition and subtraction of fractions:

If two or more fractions in a sum or difference have the same denominator, the denominator of the final result is the same, and the numerator is the sum or the difference of the original numerators.
If the fractions have different denominators, they must be brought under one denominator. This can be achieved by taking the product of the denominators of the two fractions as the new denominator and taking the product of the old numerator and the denominator of the other fraction as the new numerator for each fraction.

The corresponding formulas for the case of fractions #\frac{a}{b}# and #\frac{c}{d}#, where #a#, #b#, #c#, and #d# are polynomials, are not different from the formulas for rational numbers in Addition and subtraction of fractions:

\[ \begin{array}{lrcl}\text{same denominator}:\ \ &\frac{a}{b}+\frac{c}{b} &=& \frac{a+b}{b}\\ \text{in general: }&\frac{a}{b}+\frac{c}{d} &=& \frac{a\cdot d+b\cdot c}{b\cdot d} \end {array}\]

Again, it is not necessary to take the product of the two denominators if you can find a lower common multiple of both denominators. In that sense, the case of equal denominators is an example: the original denominator is already the common denominator.

The two fractions are real numbers for all #b\ne0# and #d\ne0#, and so is the sum of the two. But after simplification a condition for the original denominator may have been eliminated.

Put over a common denominator and expand all brackets from the expression #\frac{3}{u-5}+\frac{1}{u+5}# :

#\frac{3}{u-5}+\frac{1}{u+5}=# #\frac{4\cdot u+10}{u^2-25}#

Explanation:
\[\begin{array}{rcl}\displaystyle\frac{3}{u-5}+\frac{1}{u+5}&=&\displaystyle\frac{3(u+5)}{(u-5)\cdot(u+5)}+\frac{u-5}{(u-5)\cdot(u+5)}\\&&\phantom{ssppaacce}\color{blue}{\text{put over common denominators}}\\&=&\displaystyle\frac{3(u+5)+u-5}{(u-5)\cdot(u+5)}\\&&\phantom{ssppaacce}\color{blue}{\text{fractions with common denominators added}}\\&=&\displaystyle\frac{3u+15 + u-5}{u^2-5^2}\\&&\phantom{ssppaacce}\color{blue}{\text{brackets expanded}}\\&=&\displaystyle\frac{4\cdot u+10}{u^2-25}\\&&\phantom{ssppaacce}\color{blue}{\text{simplified}}\end{array}\]

New example

The reverse process can be tricky.

Fraction decomposition

The expanding of a fraction as a sum of fractions with denominators which divide the original denominator is called partial fraction decomposition.

The terms of the partial fraction decomposition are the partial fractions.

For example, how would you determine that #\frac{1}{x(x+1)}# is the difference of #\frac{1}{x}# and #\frac{1}{x+1}#? Partial fraction decomposition is easy if we restrict ourselves to the case of same denominators.

For the general case with a single variable, good methods are available. We will not discuss these here.

Split \({{u^2+1}\over{u}}\) in fractions with only one term in the numerator.

\({{u^2+1}\over{u}}=u+{{1}\over{u}}\)

For, #{{u^2+1}\over{u}}=\frac{u^2}{u}+\frac{1}{u} = u+{{1}\over{u}}#.

New example