Adding and Subtracting Fractions
Addition and subtraction of algebraic fractions works
basically the same way as does addition and subtraction of simple
numerical fractions:
(i.) rewrite all of the fractions as mathematically equivalent
fractions, but having the same denominator (ideally, the simplest
such common denominator).
(ii.) combine the numerators of these fractions as indicated
(adding and subtracting), retaining the common denominator.
When we work with algebraic fractions, there is then often a
third very important step:
(iii.) simplify the resulting fraction as much as possible. It
is always necessary to make sure that once the
addition/subtraction step is completed, you make a reasonable
effort to simplify the result. Usually this simplification step
will be made very much easier if you take care in step (i.) to
make sure that you have used the simplest possible common
denominator. We will illustrate the process for determining that
simplest possible denominator when the fractions to be
added/subtracted are algebraic fractions. You will see that the
strategy is essentially the same as the strategy used to
determine the least common denominator when adding or subtracting
numerical fractions, except that now symbolic factors may occur
as well as numerical factors.
Example 1:
Simplify
solution:
Factoring the two denominators gives:
2y + 2 = 2(y + 1)
and
3y + 3 = 3(y + 1)
so the simplest common denominator here is 3 Â· 2 Â· (y + 1) =
6(y + 1). Proceeding, we get
The numerator and denominator of this last fraction are fully
factored, and there are no common factors between them to cancel.
Therefore this last fraction must be the required final answer
for this problem. Even if you could see that there was little
possibility of cancellable common factors arising, it is
desirable to write the final answer in this form with numerator
and denominator fully factored.
Example 2:
Carry out the subtraction:
solution:
The denominators cannot be factored further here, so the
simplest common denominator for these two fractions is their
product, (2x + 5)(x + 2). We get
Since there appears to be no way to factor this numerator,
this last fraction is the final answer to the original
subtraction.
Example 3:
Carry out the addition:
solution:
Again, the denominators cannot be factored further, so the
simplest common denominator here will be their product, (y –
1)(y + 1). Proceeding with the general strategy, we get
as the final, completely simplified answer.
Example 4:
Perform the subtraction:
solution: The denominators here are very
simple – since they are prime numbers, the simplest common
denominator is just their product, 35. However, we must take some
care with the brackets and minus signs.
Notice particularly the use of the square brackets to ensure
that the multiplication of the numerator of the second fraction
by 7 really does include the entire numerator. Now that we have a
single fraction, we can concentrate on simplifying the numerator.
Using the principle of expanding nested brackets from innermost
to outermost, we get
This cannot be factored (since the only factors in 20 are 2,
4, 5, and 10, and it is easy to see that none of these are
factors of 163). So, our final answer here is
