Question

4/x+3 - x/x-3 - 18/x^2-9

Answer 1

add or subtract

Answer 2

hint
the common denominator for \(x+3,x-3,x^2-9\) is \(x^2-9\)

Answer 3

so i would mutiply x62-9 to all the number on the top

Answer 4

no

Answer 5

the first you would multiply top and bottom by \(x-3\)
the second by \(x+3\) and the third you would leave alone since it is already over \(x^2-9\)

Answer 6

4x+12/x2-9 - x^2-3x/x^2-9-18/x^2-9

Answer 7

not quite

Answer 8

you got them backwards

Answer 9

\[\frac{4}{x+3}=\frac{4(x-3)}{(x+3)(x-3)}\]

Answer 10

\[\frac{x}{x-3}=\frac{x(x+3)}{(x+3)(x-3)}\]

Answer 11

would i cross out the (x-3)

Answer 12

oh no!!

Answer 13

+ i mean

Answer 14

you are builing up the fractions so they have the same denominator, just like with numbers

Answer 15

*building

Answer 16

yeah im stuck

Answer 17

-x2−x+30/(x+3)(x−3) my friend got this but i dont know how he did it

Answer 18

@sweetburger

Answer 19

factor out a negative in the numerator then proceed to factor the rest of the numerator

Answer 20

actually if thats the last step u have i dont think anything else cancels you might want someone to confirm

Answer 21

ok thank you

Answer 22

@campbell_st

Answer 23

well if you go back and read what has been posted... about using common denominators

the problem is

\[\frac{4(x -3)}{(x -3)(x+3)} - \frac{x(x +3)}{(x -3)(x+3) }- \frac{18}{(x -3)(x+3)}\]

now just distribute and collect like terms in the numerator

the see what you have

Answer 24

so i dont cross out anything