jamie_23 said:
ok..the way i thought was taking the negative reciprocal of the denominator..
making it (4-1/x)(4-1/x)
---------------
(4+1/x)(4-1/x)
=16-8 1/x+1/x^2
----------------
16-1/x^2
does that work?
I haven't been taught how the clear the fractions.
Simplifying should never change the outcome of an equation. You can therefore check if your simplification is correct by letting x equal any number. The original and simplified equations will produce the same answer if the simplification is correct.
So, let x=2
Your first equation gives (4 - 1/2) / (4 + 1/2) = .77778
Your simplified equation gives (16 - 8/2 + 1/4) / (16 - 1/4) = .77778
So your simplification is correct however it is probably not a "simplification". It looks more complex. What Integral is saying is that you can get rid of a fraction by multiplying it (and all other parts of the equaion by the denomonator of the fraction. This is particularly usefull if you have the same problem in multiple parts fo the equation.
For example, (1 + 1/x) / (2 + 3/x) can be simplified by multiplying both the top and bottom lines by x (the denomonator of the fraction).
You then get (x + 1) / (2x + 3) - much simpler form than the original.
You can then multiply by the reciprical if you wish (gives you (-2x^2 + x + 3) / (-4x^2 + 9) ) but it is not as simple is it?
For the same reason as multiplying by the reciprical, multiplying both parts by the denomonator does not change the outcome of the equation because x/x = 1