Question

Help please and thank you. (This is pretty long and I would need help with the steps I am missing) ty!

Answer 1

Simplify.

\[\frac{ x+1 }{ x^2+x-12 }-\frac{ x-3 }{ x^2+7x+12 }\]

Answer 2

x^2+x-12 is (x-3) (x+4) and x^2+7x+12 is (x+3)(x+4)

Answer 3

so far so good

Answer 4

After this I need help

Answer 5

to clean it up, let a = (x-3)
b = (x+4)
c = (x+3)
\[\frac{ x+1 }{ ab }-\frac{ x-3 }{bc}\]
how do we add fractions?

Answer 6

LCM is next but I'm lost

Answer 7

I need help understanding how to get the LCM in this equation

Answer 8

\[\frac cc\frac{ x+1 }{ ab }-\frac aa\frac{ x-3 }{bc}\]
\[\frac{ c(x+1)-a(x-3) }{ abc }\]
but we defined x-3 as a
\[\frac{ c(x+1)-aa }{ abc }\]
\[\frac{ c(x+1)-a^2 }{ abc }\]
do you have to expand it all out?

Answer 9

ah, the lcm would be the product of the denominators. abc

Answer 10

Using this method of cleaning it up I am unable to grasp

Answer 11

do we know the value of (x-3) ?

Answer 12

My answer was 10x-21/(x+4)(x+3)(x-3) but that is incorrect

Answer 13

the bottom looks good

lets work on the top

Answer 14

-2+12/(x+4)(x+3)(x-3) is incorrect as well that was my second attempt

Answer 15

c(x+1) - a^2

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

(x^2+x+3x+3) - (x^2-6x+9)

x^2 +4x +3
-x^2 +6x - 9
-------------
0 +10x-6

Answer 16

meant to say -2x+12**/"()()()"

Answer 17

yeah, the top works out to be 10x-6

Answer 18

by clean up i meant that we keep the same terms, but just dont clutter the space with all that (x+..)(x-....) typing.

its the objects the get moved about so what we name them is not important was all

Answer 19

I understand I just don't follow that and what you wrote before where you get 0+10x-6 I am trying to follow how you go there the process

Answer 20

there was also a part you had aa one is a and the other a represents x-3
I'm just completely lost

Answer 21

since i defined a as (x-3)

i got to a point of a(x-3), which is just a, times a ... a^2. or (x-3)^2

Answer 22

spose we dont clean it up ....

Answer 23

theres jsut more room for typing in an error is all ...
\[\frac{ x+1 }{ x^2+x-12 }-\frac{ x-3 }{ x^2+7x+12 }\]
\[\frac{ (x+1) }{(x-3)(x+4) }-\frac{ (x-3) }{(x+3)(x+4) }\]
\[\frac{ (x+1)(x+3) }{(x-3)(x+3)(x+4) }-\frac{ (x-3)(x-3) }{(x-3)(x+3)(x+4) }\]