Question

add and simplify. (-3x/x^2-9) + (2/x-3)

Answer 1

\[\frac{ -3x }{ x^2-9 }+\frac{ 2 }{x-3 }\]

To add you'll need an LCD, but to find an LCD you have to make sure all denominators are factored so that you can see what factors there are. Are both of these denominators fully factored?

Answer 2

\[\frac{ -3x }{ (x-3)(x+3) }+\frac{ 2 }{ (x-3) }\]

Answer 3

can u cross multiply after they are factored?

Answer 4

Cross-multiplying has to do with solving an EQUATION. here you don't have an equation, just an EXPRESSION to evaluate. You need a common denominator. Do you know what it is?

Answer 5

(x-3)

Answer 6

?

Answer 7

Well, what are the denominators? Another user factored the first denominator for you above (I wanted you to do it yourself, lol :) so you have denominators:

(x-3)(x+3)
and (x-3)

The LCD MUST BE a MULTIPLE of ALL the denominators. That is you, must be able to multiply each denominator by SOMETHING (maybe 1) and GET the LCD. So need:

(x-3)(x+3)(?)=LCD
(x-3)(?)=LCD

Now, you said LCD=x-3.... but then how would you get that first denominator?
(x-3)(x+3)(?)=(x-3)

Answer 8

well, then would the two sides just equal each other?

Answer 9

I'm not sure what you mean by that. There are not "two sides" as this isn't an equation. At least not as you presented it. This is an addition problem. You are adding terms, not solving an equation.

It's like the difference between me asking you:
1. ADD: 2+7-5
vs.
2. SOLVE: 2x+7=5

One is addition. The other is an equation.

(This is a common confusion I see in algebra students, but it's important that you wrap your head around the difference.)

Answer 10

oh I see

Answer 11

i still don't see how to get the LCD

Answer 12

the simplest form is \[\frac{ -x+6 }{ x ^{2}-9 }\]

Answer 13

OK.... the LCD must be a multiple of the all the denominators, so you need to look at the factors in each den'r and make sure you have each of those factors in the LCD.

E.g., for NUMBERS, suppose you had denominators of 6 and 9.
6=2(3) and 9=3(3)
So the LCD MUST HAVE one factor of 2, and two factors of 3, eg. LCD=2(3)(3)=18

Make sense so far?

Answer 14

yes, that makes sense

Answer 15

Now here, your terms have algebraic expressions in the denominator - not just numbers. One has:
(x-3)
so you have to have that as a factor of the LCD.

The other has (x-3)(x+3)
Now, the (x-3) is already "covered", but you also need a factor of (x+3). So your LCD is:

(x-3)(x+3)

Answer 16

both of them?

Answer 17

Now build up the the 2nd term over the LCD (no need to do anything to the first term in this case, since it already HAS the LCD):

\[\frac{ -3x }{(x-3)(x+3) }+\frac{ 2 }{(x-3) }\times \frac{ (x+3)}{(x+3) }\]

Answer 18

Yes, both of them. Their product.

Answer 19

Hannah, it's no different than with numbers. ONE of your den'rs is already a multiple of the other, so that denominator will BE your LCD.

Like if you have a denominator of 5 in one fraction and 10 in the other, then your LCD=10, right?

It's the same idea here. It doesn't always work out that way (that the LCD IS one of the den'rs in the problem), but here it does.

Answer 20

\[-3x/(x-3)(x+3)+2x+6/(x-3)(x+3)\]

Answer 21

oh!!!! ok

Answer 22

i think i was just confused because of the expressions

Answer 23

Need some parens () there, but I know what you mean, so yes. :)
\[\frac{ -3x }{(x-3)(x+3) }+\frac{ 2x+6 }{(x-3)(x+3) }\]

Answer 24

OK, now what? You have both terms with a common denominator, so you know what to do next, right?....


Add the numerators (combine like terms), and put it all over the LCD, so you'll combine into 1 term.

Answer 25

Confusion is OK, as long as you work through it. Sometimes that's when you learn the best! :)

Answer 26

so when you add them you would get \[-x+6/x^2-9\]

Answer 27

and you can't simplify it more than that

Answer 28

I think you've got it!! :) Well done!

Answer 29

thank for your help!

Answer 30

you're welcome, glad to help. :)