Question 1168889: Did I add these 2 equations correctly below?
3a
a+4
+
2a2−16a
a2−16
=
5a3−8a2−112a
a3+4a2−16a−64
=
a(5a−28)(a+4)
(a+4)(a+4)(a−4)
=
5a2−28a
a2−16
Thank you Answer by MathTherapy(10551) (Show Source): You can put this solution on YOUR website! Did I add these 2 equations correctly below?
3a
a+4
+
2a2−16a
a2−16
=
5a3−8a2−112a
a3+4a2−16a−64
=
a(5a−28)(a+4)
(a+4)(a+4)(a−4)
=
5a2−28a
a2−16
Thank you
Again, you need to PRESENT these problems better! This one should be written like this: 3a/(a + 4) + (2a^2 - 16a)/(a^2 - 16)
Your final answer is CORRECT, considering that what you have is: