Alegrabic Fraction Simplification

[thomas49th]

[SOLVED] Alegrabic Fraction Simplification

Homework Statement

Show that:
[tex]\frac{3x}{x+1} - \frac{x+7}{x^{2}-1}, x > 1[/tex]

can be written as:

[tex] 3 - \frac{4}{x-1}[/tex]

The Attempt at a Solution

Well i can see the difference of 2 squares on the bottom of the second fraction

[tex]\frac{3x}{x+1} - \frac{x+7}{(x+1)(x-1)}[/tex]

cross multiply and x+1 cancels out

giving

[tex]\frac{3x(x-1)-(x+7)}{(x+1)(x-1)}[/tex]

the top factorises to (3x-7)(x+1) cancelling the (x+1)

giving me
[tex] \frac{3x-7}{x-1}[/tex]

But that doesn't equate to [tex] 3 - \frac{4}{x-1}[/tex] does it?

Where have i gone wrong
Thanks :)

 

[Defennder]

You are correct actually. Just perform polynomial long division on your second last expression and you'll get the answer.

Another way you could get it from the original question would be to do polynomial long division on the left term and breaking the one on the right down to partial fractions, then canceling common factors.

 

[GregA]

Defennder is absolutely correct, though a little trick you can also employ in these situations is the following:
[tex] \frac{3x -7}{x-1} = \frac{3x-3-4}{x-1} = \frac{3(x-1)}{x-1} - \frac{4}{x-1} = 3-\frac{4}{x-1}[/tex]

 

[thomas49th]

Ahh cheers :) That's pretty cool. Yeh should of spotted the 3 and 4 and 7 relationship. Cheers :)