Question

((x-2)/(x-3))-((x+1)(x+3))+((x-9(/x^2-9))

Answer 1

I think the best way to do this is to just multiply out the first two terms, and for the third term, factor the denominator to (x-3)(x+3).

Answer 2

can you multiply it? I dont understand this. it looks like subtraction to me

Answer 3

oh sorry, I didn't see the / in the first term.
Ok, for the second term, (x+1)(x+3), multiply this out
Then for the third term, (x-9)/(x^2-9), factor the denominator

Answer 4

uh oh i typed this out wrong its supp((x-2)/(x-3))-((x+1)/(x+3))+((x-9)/(x^2-9))osed to be

Answer 5

\[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2 -9}\]
Recognize that \(x^2 - 9 = (x+3)(x-3)\) and put them all under a common denominator of \(x^2-9\)

Answer 6

Over a common denominator rather.

Answer 7

so i need to multiply the first two fractions by (x+3) or x-3?

Answer 8

Multiply the first by (x+3) on top and bottom, and the second by x-3 on top and bottom.

Answer 9

Oh, was the second term also a fraction?
Remember when you learned how to add fractions with different denominators? Use the same logic. Multiply the first fraction by (x+3)/(x+3) and the second by (x-3)/(x-3) so that they all have a common denominator.
Basically what polpak is saying but it is worth noting that this is just like what you do when you are adding normal fractions with different denominators.

Answer 10

Oh. I may have mistyped it. Was the middle term a fraction?

If not it'd be
\[\frac{x-2}{x-3} +(x+1)(x+3) +\frac{x-9}{x^2-9}\]

Answer 11

But you still want to get them over a common denominator of \(x^2-9\)

Answer 12

polpak, she posted that they were all fractions a few minutes ago. I missed it too.

Answer 13

you were correct the first time i mistyped it... thank you both for your help! its greatly appreciated

Answer 14

so i came up with 0(x+3)(x-3) is this correct? i somehow dont think it is

Answer 15

im sorry 0/(x-3)(x+3)

Answer 16

I got something else. Did you forget that the second fraction is being subtracted? I almost forgot that.

Answer 17

no i subtracted it but it all seemed to cancel out. what am i doing wrong?

Answer 18

\[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2-9} \]
\[= \frac{(x-2)(x+3)}{x^2-9} - \frac{(x+1)(x-3)}{x^2-9} + \frac{x-9}{x^2-9} \]
\[= \frac{[x^2+x-6] - [x^2-2x -3] + [x-9]}{x^2-9} \]
\[= \frac{x^2-x^2+x + 2x + x -6+3 -9}{x^2-9} \]
\[= \frac{4x-12}{x^2-9} \]
\[= \frac{4(x-3)}{x^2-9} \]
\[= \frac{4(x-3)}{(x+3)(x-3)} \]
\[= \frac{4}{x+3} \]

Answer 19

Where'd you go wrong?

Answer 20

And hopefully you can follow that mess.

Answer 21

i didnt change the signs in the second fraction AND i can follow it

Answer 22

Thank you so much!

Answer 23

Cool, np =)

Answer 24

4/( x+3)