In this lesson we are going to simplify a relatively simple algebraic fraction.

First, let me explain the syntax we will be using. All algebraic expressions will be written in 'one-line' notation, using common calculator operations(+,-,*,/ and ^). The only one that you might not be familiar with, could be '^' which stands for exponentiation (i.e. x^2 means "x squared"). In addition, when radicals are needed, "R" will be used (i.e. 3R(x^2) would stand for "cube root of x squared"). This syntax is a little cumbersome; however you can cut and paste any of these expressions to a 'syntax converter' located at this algebra resources page. It will create a nice looking textbook notation.

So, here is the problem:

(4*x^2-9*y^2)/(2*x+3*y)^2

The first thing that has to be done is completely factor both numerator and the denominator. Since denominator is already factored, let's concentrate on the numerator. If you look at it closely, you will realize that it represents a difference of two squares. We first need to make this difference explicit (more obvious):

((2*x)^2-(3*y)^2)/(2*x+3*y)^2

Now it is a little clearer that we can use the following 'difference of squares' factoring formula:

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

In our case, a is equal to 2x and b is equal to 3y. The application of the above formula will result in:

(2*x-3*y)*(2*x+3*y)/(2*x+3*y)^2

At this point you should notice that one of the factors is the same in both numerator and denominator. That means that we can reduce them:

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

What exactly does (2-1) mean in the above expression? Well, the implied power of (2x-3y) in the numerator was "1" - and the reduction rule says that we need to subtract the smaller power from the larger.

Finally, we end up with the following simplified fraction:

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

This wasn't too hard, was it?

Stay tuned for more algebra lectures!