# Algebra Intro 12: Adding and Subtracting Fractions

Once someone knows how to multiply fractions, and is comfortable creating equivalent fractions by multiplying by a fraction that equals 1, they have to tools needed to add and subtract fractions.

### Why Can’t I Just Add Two Fractions As Written?

Consider the fraction “two thirds.” The phrase as written can be represented in two ways: $\dfrac{2}{3}$

or $2\cdot\dfrac{1}{3}$

The first version is what most people probably think of when they read or hear “two thirds”.  However, the second version is also described perfectly by the phrase, as it represents “two” of the quantity known as a “third” – just like we might say “two apples”. By thinking of fractions in this second way, it is easier to understand when fractions can be added as written, and when they cannot.

What if someone asks you to add two apples and three oranges, or two meters and three pumpkins? You cannot give them an answer because you are being asked to combine dissimilar things: you are being asked to add quantities that do not have the same units. You could finesse the apples and oranges question by referring to both as “pieces of fruit”, by converting the units you were given to a common unit. However, this approach does not work so well with meters and pumpkins.

Addition and subtraction can only be carried out when both quantities either have no units (they are abstract numbers) or have the same unit (they are quantities of the same thing). This is true for the addition or subtraction of fractions as well.

The problem $\dfrac{2}{3}+\dfrac{3}{4}$

can be rewritten using the approach described above as $2\cdot\dfrac{1}{3}+3\cdot\dfrac{1}{4}$

which means you are being asked to add two “thirds” to three “quarters”.  The two quantities do not have the same units, and thus cannot be added as written. However, using the rules of algebra, you can convert both fractions to equivalent fractions with the same units – just as you might have converted apples and oranges to “pieces of fruit” in the example above: $\dfrac{2}{3}+\dfrac{3}{4}$ $=\dfrac{2}{3}\cdot\left(\dfrac{4}{4}\right)+\dfrac{3}{4}\cdot\left(\dfrac{3}{3}\right)$ $=\dfrac{2\cdot 4}{3\cdot 4}+\dfrac{3\cdot 3}{4\cdot 3}$ $=\dfrac{8}{12}+\dfrac{9}{12}$ $=8\cdot\dfrac{1}{12}+9\cdot\dfrac{1}{12}$

You now have two quantities of the same unit, “twelfths”.  You also have a sum of two expressions with a common factor: one twelfth.  If that common factor is factored out (un-distributed), the problem becomes $(8+9)\cdot\dfrac{1}{12}$ $=17\cdot\dfrac{1}{12}$ $=\dfrac{17}{12}$

All of which explains the rules you were taught in elementary school, which provide a minor shortcut to the process used above.  To add or subtract two fractions, first multiply each by some form of the number one to create a common denominator for both fractions.  Once both have a common denominator (changing one or both numerators in the process) you are welcome to combine the numerators.

### Lowest Common Denominator

If a problem asks you to add or subtract two fractions that do not have the same denominator, you need to find a common denominator for them.  It is often more efficient if that denominator is the “lowest common denominator”, or “least common multiple”: the smallest number that is evenly divisible by both denominators.

One way to find the lowest common denominator is to rewrite both denominators as a products of their prime factors.  For example, $\dfrac{1}{30}+\dfrac{1}{70}$ $=\dfrac{1}{2\cdot 3\cdot 5}+\dfrac{1}{2\cdot 5\cdot 7}$

Once written this way, group all factors that are common to both denominators together: $=\dfrac{1}{(2\cdot 5)\cdot }+\dfrac{1}{(2\cdot 5)\cdot }$

The remaining factors, the ones in square brackets above, are the ones that are missing from the other denominator, and are therefore the ones I will need to use when creating an equivalent fraction: $=\dfrac{1}{(2\cdot 5)\cdot }\cdot\left(\dfrac{7}{7}\right)+\dfrac{1}{(2\cdot 5)\cdot }\cdot\left(\dfrac{3}{3}\right)$ $=\dfrac{7}{2\cdot 3\cdot 5\cdot 7}+\dfrac{3}{2\cdot 3\cdot 5\cdot 7}$ $=\dfrac{10}{2\cdot 3\cdot 5\cdot 7}=\dfrac{10}{210}=\dfrac{1}{21}$

A slight shortcut to the above approach was described here by fellow blogger Josh Rappaport.  If you create a fraction using your two denominators, then simplify it $\dfrac{30}{70}=\dfrac{3}{7}$

the process of simplifying the fraction will cancel out all common factors (the ones I had in parentheses above), and leave behind the product of the unique factors (the ones I had in square brackets above).

The 3 in the numerator came from the original denominator of 30, so I need to apply it to the 70 to produce the lowest common multiple.  The  7 came from the 70, so I need to apply it to the 30. This exactly what was done using the first approach I described.

### Algebraic Fractions

The rules for working with fractions do not change when variables are introduced. The problem $\dfrac{2a}{b}-\dfrac{3c}{d}$

asks us to subtract two fractions that do not have a common denominator. Therefore our first step has to be to seek a common denominator, to get both expressions “using the same units”. This can always be done by multiplying by one, in the form of the other fraction’s denominator over itself… however this may not produce the “lowest” common denominator. $\dfrac{2a}{b}\cdot\left(\dfrac{d}{d}\right)-\dfrac{3c}{d}\cdot\left(\dfrac{b}{b}\right)$ $=\dfrac{2ad}{bd}-\dfrac{3bc}{bd}$ $=\dfrac{2ad-3bc}{bd}$

In this case, multiplying by the other’s denominator did produce the “lowest common denominator”, since the numerator and denominator in the result above do not have any common factors.

By carrying out the subtraction of the two algebraic fractions in the original problem, we have combined them into a single fraction. Algebra is all about changing the appearance of an expression without changing the quantity that it represents – so if you prefer to work with one fraction instead of two at some point in a problem, you now have the tools to combine the two.

Similarly, when presented with a fraction that has multiple terms in its numerator (separated by plus or minus signs), you are welcome to reverse the above process and break it up into multiple fractions if that helps to solve the problem: $\dfrac{a+b-c}{d}$ $=\dfrac{a}{d}+\dfrac{b}{d}-\dfrac{c}{d}$ 