1. If the unknown is in the denominator #q(x)# of such a fraction, we multiply both sides of the equation by that denominator #q(x)# to bring the unknown "above the line."
2. The result is a new equation, which is only equivalent to the old one if #q(x)\ne0#. If we solve the new equation, we find the solution to the old one by selecting the solutions #x# for which #q(x)\ne0#.

Example

The function #f(x) = x^{2}\cdot (x-1)^{-1}# is a quotient function, because it can be written as #\frac{x^2}{x-1}#. It is defined everywhere except in #x=1#.

To find out where this function takes the value #4#, we solve the equation #f(x) = 4#.

\[\begin{array}{rcl} f(x) &=& 4\\ &&\phantom{xx}\color{blue}{\text{the equation to solve}}\\ \dfrac{x^2}{x-1} &=& 4 \\ &&\phantom{xx}\color{blue}{\text{substituted function rule}}\\{x^2} &=& 4\cdot (x-1) \\&&\phantom{xx}\color{blue}{\text{eliminated denominator}}\\ x^2-4x+4 &=&0 \\ &&\phantom{xx}\color{blue}{\text{all terms to the left-hand side}}\\ x &=&2\\ &&\phantom{xx}\color{blue}{\text{solved quadratic equation}}\\ \end{array}\] We verify that the denominator of #f(x)# has the value #2-1=1# for #x=2#, so that the solution #x=2# is indeed also a solution of #f(x) = 4#. It is clear that #x=2# is the only value in which #f(x)# has the value #4#.

Calculation rules for quotient functions

The following well-known calculation rules for fractions also apply if #p#, #q#, #r# and #s# are functions. Obviously, all denominators must be unequal to #0#.

1. #\displaystyle \dfrac{p}{q}+\dfrac{r}{s} = \dfrac{s\cdot p+q\cdot r}{q\cdot s}#
2. #\displaystyle \dfrac{p}{q}+\dfrac{r}{q} = \dfrac{ p+ r}{ q}#
3. #\displaystyle \dfrac{r\cdot p}{r\cdot q}= \dfrac{p}{q}#
4. #\displaystyle r\cdot \dfrac{p}{q}= \dfrac{r\cdot p}{q}#
5. #\displaystyle\dfrac{ \dfrac{p}{q}}{ \dfrac{r}{s}}= \dfrac{s\cdot p}{q\cdot r}#

The values of #x# in which the functions are defined are not necessarily the same for the left-hand- and right-hand side of the equation. For example, the function #\frac{x}{x}# is not defined in #0#, but the simplified fraction #1# (obtained by applying rule 3 with #r=x# and #p=q=1# ) is.