In this lecture you will learn how to add three algebraic fractions with different denominators.

The ‘one-line’ mathematical expression syntax that we have to use here is a little cumbersome, so if you want it converted to a pretty-looking ‘text book’ notation, you can do so by using the conversion tool available on this algebra resources page.

We will solve the following problem:

2/(x-2)-3/(x+2)+1/(x^2-4)

As you can see, there are three fractions here. You can also see that the denominators are all different (duuh!). The first step in adding fractions with different denominators consists of finding the Least Common Denominator (LCD). In order to be able to do that, we must have all denominators factored. There is nothing to factor in the first two, so that leaves the third one:

2/(x-2)-3/(x+2)+1/((x-2)*(x+2))

We have employed the ‘difference of two squares’ formula:

a^2-b^2 = (a-b)*(a+b).

In this problem “a” was equal to “x” and “b” was equal to “2” (because 2 squared equals 4).

Now it should be pretty obvious what the LCD is – it has to contain all the factors found in all denominators – and that would be this expression: (x-2)*(x+2).

Once we know the least common denominator, we create the numerator in the following way:

- divide LCD with each fraction’s denominator and multiply the result with the corresponding numerator. Add all these terms and voila, here is your new numerator:

((x+2)*2+(x-2)*(-3)+1)/((x-2)*(x+2))

The rest of the process is easy – we just need to simplify the resulting numerator. First we multiply out these two terms: 2(x+2) and (-3)(x-2); don’t forget the ‘-‘ sign in front of number 3!

((2*x+2*2)+(-3*x-3*(-2))+1)/((x-2)*(x+2))

Then we get rid of the parentheses:

(2*x+4-3*x+6+1)/((x-2)*(x+2))

And finally add the like terms:

(-x+11)/((x-2)*(x+2))

At this point we are done (keep in mind that in some more complex problems, you will need to factor the new numerator and try to reduce it with the denominator). We leave the denominator factored, as that form is usually considered simpler.

So, how hard was this ?

A few more lessons, and you will be on your way to better algebra grades. . .

John Madigan is the principal content developer for Algebra Tutor , a software program that produces solution steps and explanations for algebra problems such as the one above. In his spare time he tutors math to those that just “can't get it".

*November 06, 2005*(471)