Subtract Fractions (unlike denominators)

Subtracting fractions is just like adding fractions (except, of course, you are taking away instead of adding!).

Remember, when you subtract fractions, you need to make the fractions have the same denominator (which means that the pieces you are subtracting are the same size -- for more info on finding common denominators, review the Add Fractions Lesson).

Like when we add fractions, we think that when you subtract fractions, it's easiest if you write them vertically.  Vertically, you have lots of room to figure out and write your new, equivalent fractions that have the same denominator.

So, if you are given the problem $\dfrac{ 1}{5} - \dfrac{ 2}{15}$, write it like this:

$$\eqalign{\text{}\dfrac{1\;}{5\;}&\\\underline{-\dfrac{2}{15}}&}$$

Then, think about what common denominator $\dfrac{1}{5}$ and $\dfrac{2}{15}$ could share.  A common denominator is a number that both denominators can go into evenly (a multiple of both numbers).  So, what do $5$ and $15$ both go into evenly?  They both go into $15$ evenly.  So, convert both fractions to fractions with the denominator of 15:

$$\eqalign{\dfrac{1\;}{5\;}-\dfrac{2}{15}\\ \text{Write it vertically  }&\left\{ \begin{array}{rcl} \dfrac{1\;}{5\;}&\\ -\dfrac{4}{15}& \\\hline \qquad \qquad\\ \qquad  \end{array}\right.  \\\text{Find the common denominator: 15  }&\left\{ \begin{array}{rcl}  \dfrac{1\;}{5\;}&=&&\dfrac{?}{15}\\-\dfrac{4}{15}&=&&\dfrac{?}{15}\\\hline \qquad \qquad \\ \qquad \end{array}\right.\\\begin{array}{rcl}\text{Multiply each fraction by a fractional version of 1  }\\\text{that will make each denominator equal 15  }\end{array}&\left\{ \begin{array}{rcl}\dfrac{1}{5}&\times&&\dfrac{3}{3}=&&&\dfrac{3}{15}\\-\dfrac{4}{15}&\times&&\dfrac{1}{1}=&&&\dfrac{4}{15}\\ \hline \qquad \qquad \end{array}\right.\\\text{Subtract the fractions with common denominators  }&=\begin{array}&\quad && \quad &&& \quad &&&&\quad\dfrac{7}{15}\end {array}}$$

Sometimes, it's not quite as easy to find a common denominator.  If the smaller number goes evenly into the larger denominator, the larger denominator is the perfect common denominator.  But, other times, you have to go with a number that is bigger than either denominator.

$$\eqalign{&\dfrac{6}{7}-\dfrac{5}{9}\\\text{Write it vertically  }&\left\{ \begin{array}{rcl}\dfrac{6}{7}&\\-\dfrac{5}{9}&\\ \hline \qquad \qquad \\ \qquad \end{array}\right.\\\text{Find the common denominator: 63  }&\left\{ \begin{array}{rcl}\dfrac{6}{7}&=&&\dfrac{?}{63}\\-\dfrac{5}{9}&=&&\dfrac{?}{63} \\\hline \qquad \qquad\\ \qquad \end{array}\right.\\ \begin{array}{rcl}\text{Multiply each fraction by a fractional version of 1  }\\\text{that will make each denominator equal 63  } \end{array} &\left\{ \begin{array}{rcl}\dfrac{6}{7}&\dfrac{\times}{\times}&&\dfrac{9}{9}=&&&\dfrac{54}{63}\\-\dfrac{5}{9}&\times&&\dfrac{7}{7}=&&&\underline{\dfrac{35}{63}}\\\hline \qquad \qquad\\ \qquad \end{array}\right. \\\text{Subtract the fractions with common denominators  }&= \begin {array} & \quad && \quad &&& \quad &&&& \quad \dfrac{19}{63}\end{array}}$$

Othertimes, you will end up with an answer that you have to reduce or simplify.

$$\eqalign{\dfrac{5\;}{8\;}-\dfrac{3}{12}\\\text{Write it vertically  } &\left\{ \begin{array}{rcl} \dfrac{5\;}{8\;}&\\-\dfrac{3}{12}& \\\hline \qquad \qquad\\ \qquad \end{array}\right.\\\text{Find the common denominator: 24  }&\left\{ \begin{array}{rcl} \dfrac{5}{8}&=&&\dfrac{?}{24}\\-\dfrac{3}{12}&=&&\dfrac{?}{24} \\\hline \qquad \qquad\\ \qquad \end{array}\right.\\ \begin{array}{rcl} \text{Multiply each fraction by a fractional version of 1  }\\\text{that will make each denominator equal 24  }\end{array}&\left\{ \begin{array}{rcl}\dfrac{5}{8}&\times&&\dfrac{3}{3}=&&&\dfrac{15}{24}\\-\dfrac{3}{12}&\times&&\dfrac{2}{2}=&&&\dfrac{6}{24} \\\hline \qquad \qquad\\ \qquad \end{array}\right.\\\text{Subtract the fractions with common denominators  }&= \begin {array} & \quad && \quad &&& \quad &&&& \quad\dfrac{9}{24}\end{array}\\\text{Reduce answer by dividing by common factor}=&\begin {array} & \quad && \quad &&& \quad &&&& \quad\dfrac{9}{24}\dfrac{\div}{\div}\dfrac{3}{3}=\dfrac{3}{8}\end{array}}$$

Overall, once you have fractions with common denominators, you just have to subtract the numerators, keep the denomintors, and simplify if necessary.