SOLUTION: Did I also subtract the below equations correctly? 3a a+4 − (2a2−16a a2−16)= a3+8a2+16a a3+4a2−16a−64= a(a+4)(a+4) (a+4)(a+4)(a−4)= a a−4 Th

Algebra ->  Equations -> SOLUTION: Did I also subtract the below equations correctly? 3a a+4 − (2a2−16a a2−16)= a3+8a2+16a a3+4a2−16a−64= a(a+4)(a+4) (a+4)(a+4)(a−4)= a a−4 Th      Log On


   

Question 1168890: Did I also subtract the below equations correctly?

3a
a+4
−
(2a2−16a
a2−16)=
a3+8a2+16a
a3+4a2−16a−64=

a(a+4)(a+4)
(a+4)(a+4)(a−4)=
a
a−4
Thank you


Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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Tell exactly what you are given and what you want to know how to do or to understand.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Did I also subtract the below equations correctly?

3a
a+4
−
(2a2−16a
a2−16)=
a3+8a2+16a
a3+4a2−16a−64=

a(a+4)(a+4)
(a+4)(a+4)(a−4)=
a
a−4
Thank you
Your simplification doesn't seem to make sense.

However, your final answer is CORRECT, considering that what you have is: 3a%2F%28a+%2B+4%29+-+%282a%5E2+-+16a%29%2F%28a%5E2+-+16%29

By the way, you need to present these problems better. This one should be written like this: 3a/(a + 4) - (2a^2 - 16a)/(a^2 - 16)