Question

Please help! I've been working on this for two hours!
Simplify this expression and state any restrictions on the variables:
(x+1)/(x^2+2x-8) - (x/(4x-8)

Answer 1

@amistre64 @Agent_Sniffles @TuringTest

Answer 2

Can you factor a number out of that second term?

Answer 3

Can't you factor out a two?

Answer 4

You can factor out a four too.

Answer 5

Okay, so how do I go about doing that?

Answer 6

well factorise both denominators

\[\frac{x +1}{(x +4)(x -2)} - \frac{x}{4(x -2)}\]

you'll need a common denominator of 4(x + 4)(x -2)

so the 1st fraction is multiplied by 4/4

and the 2nd fraction is multiplied by (x +4)/(x +4)

which will give

\[\frac{4(x + 1)}{4(x +4)(x -2)} - \frac{x(x +4)}{4(x +4)(x -2)}\]

Answer 7

hehe like so! ^

Answer 8

Sorry - I didn't mean you could factor out a four AND a two. You can factor out a two or a four. A four is a better idea though

Answer 9

the restrictions are found by looking at the denominator

and solving

x + 4 = 0

x - 2 = 0

just simplify the numerator

Answer 10

so is the answer ((4x+4)/(x+4)) - ((x^2+4x)/(x-2))?

Answer 11

How did you cancel the 4(x-2) on the left term and the 4(x+4) term on the right? I don't think its possible.

Answer 12

I'm not sure.... What should the answer be?? I'm confused.. Sorry, I just need a good grade on this

Answer 13

well start by leaving the denominator alone

\[\frac{ 4x + 4 - x^2 - 4x}{4(x+4)(x -2)}\]

see if you can simplify the numerator

Answer 14

(-8x^2-x^2+4)/((4(x+4)(x-2)) ??

Answer 15

so the numerator can be written as

\[\frac{-(x^2 -4)}{4(x + 4)(x -2)}\]

the numerator can be factorised... its the difference of 2 squares...

a common factor can be removed.

but a point of discontinuity will exist...

Answer 16

ok so the answer is (-(x^2-4)/4(x+4)(x-2)) and no restrictions?

Answer 17

Try factoring the numerator. There are restrictions.

Answer 18

there are restrictions...

1 a vertical asymptote

2. a point of discontinuity