Melody? Could you please help me...

SelinaMarie:

Simplify the difference. State the restrictions.

(1/x²-1) - (2/x²+3x)

Could you please show me step by step.

hello SelinaMarie,
Why don't you sign up properly, and, of course I will help you.

(1/x²-1) - (2/x²+3x)
= 1/x²-1 -1 (2/x²+3x) You have to multiply everything in the bracket by -1
= 1/x²-1 - 2/x² - 3x
= 1/x²- 2/x² - 3x -1
= -1/x² - 3x -1

Write your fractions upright and it will help you 'see' better.
If you don't understand any of the steps let me know.
Melody.

Oh my goodness, I'm so sorry, I wrote it wrong. It was supposed to be:
(1/(x²-1)) - (2/(x²+3x))
I'm so sorry.

SelinaMarie:

Oh my goodness, I'm so sorry, I wrote it wrong. It was supposed to be:
(1/(x²-1)) - (2/(x²+3x))
I'm so sorry.

= [ (x²+3x) - 2 (x²-1) ] / [ (x²-1)(x²+3x) ]

= [ x²+3x -2x² + 2 ] / [ (x²-1)(x²+3x) ]

= [ -x² +3x + 2 ] / [ (x²-1)(x²+3x) ]

or

= [ x² - 3x - 2 ] / [ (1- x²) (x²+3x) ]

The denominator is ==> x(1-x)(1+x)(x+3)
A denominator cannot =0 therefore the restrictions are x is not equal to 0, 1,-1, or -3

are you sure you got the question right this time Selina because these questions usually fall out much better than this.