SOLUTION: m/m^2-m-12 + 5/m+3 This is what i came up with but i'm not exactly sure how i got it. I'm not sure how to get the denominators the same. Please walk me through this proble

Question 36872: m/m^2-m-12 + 5/m+3
This is what i came up with but i'm not exactly sure how i got it. I'm not sure how to get the denominators the same. Please walk me through this problem step by step. Thank you. I don't have a text book. My instructor gives us problems from his book.
5(m-4)/m+3(m+3)(m-4)
m+5m-20
answer =
6m-20/m^2-m-12
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
m/m^2-m-12 + 5/m+3
=m/[(m-4)(m+3)] + 5/(m+3)
The LCD = (m-4)(m+3)
Rewrite each fraction with this LCD is its denominator, as follows:
= m/LCD + 5(m-4)/LCD
Now add the fractions to get:
=[m+5(m-4)]/LCD
Simplify as much as possible, as follows:
=[6m-20]/[(m-4)(m+3)]
Your answer is correct.
Cheers,
Stan H.

RELATED QUESTIONS
I just want to make sure that i have done this correctly. Simplify... (answered by stanbon,praseenakos@yahoo.com)

1/square root 8 = 4^m-3 how do i solve this problem? the answer is m = 9/4 but im not... (answered by stanbon,ewatrrr)

Simplify. (m^4)^3(m^2)^9/(m^3)^2 I came up with two answers for this problem and not... (answered by jim_thompson5910,stanbon)

Hi, I'm confused about solving for this trig equation: 3cos(x)+cos^2(x)+2=sin^2(x). I (answered by ikleyn)

I need to simplifi this rational expression. The next step I think is to cancel out... (answered by kietra)

m^2+7m+12/m^2-8m+16 + m+3/m^2-16 =(m+3)(m+4)/(m-4)(m-4)+(m+3)/(m-4)(m-4)... (answered by Edwin McCravy)

Conjecture: Find all positive integers 'a' such that there exists an integer 'm' with the (answered by richard1234)

Find the numerator: (3m+6)/(m^2-m)=(?)/(m^3-5m^2+4m) I have a problem asking me to... (answered by stanbon,MathTherapy)

I am pretty sure I have this under control but I would like to verify my answers.... (answered by rwm)