SOLUTION: Divide and if possible, Simplify. (a+7)divided by 3a^2+14a-49(over)a^2+8a+7 This was my attempt but teacher said it was wrong. Can you please help me so I can see where I w

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Divide and if possible, Simplify. (a+7)divided by 3a^2+14a-49(over)a^2+8a+7 This was my attempt but teacher said it was wrong. Can you please help me so I can see where I w      Log On


   

Question 251293: Divide and if possible, Simplify.
(a+7)divided by 3a^2+14a-49(over)a^2+8a+7

This was my attempt but teacher said it was wrong. Can you please help me so I can see where I went wrong?
(a+7)/ 3a^2+14a-49 over a^2+8a+7
(a+7)/[(3a-7)(a+7)] over [(a+7)(a+1)]
Cancel(a+7) to get 1/[(3a-7)] over [(a+7)(a+1)]
Rewrite
[1/(3a-7)]/[(a+7)(a+1)]/1
Invert the denominator and multiply
= 1/[(3a-7)(a+7)(a+1)]
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
looks like you got the factors ok.

your original equation is:

%28a%2B7%29%2F%28%28%283a%5E2%2B14a-49%29%2F%28a%5E2%2B8a%2B7%29%29%29

you factored correctly to get:

%28a%2B7%29%2F%28%28%28%283a-7%29%2A%28a%2B7%29%29%2F%28%28a%2B7%29%2A%28a%2B1%29%29%29%29

you canceled correctly to get:

1%2F%28%28%283a-7%29%2F%28%28a%2B7%29%2A%28a%2B1%29%29%29%29

it looks like where you went wrong was when you inverted the denominator.

since a%2F%28b%2Fc%29 = %28a%2Ac%29%2Fb, your answer should have been:

%281%2A%28a%2B7%29%2A%28a%2B1%29%29%2F%283a-7%29 which becomes:

%28%28a%2B7%29%2A%28a%2B1%29%29%2F%283a-7%29