Question

3-(a-4/a+4)=a^2-8/a+4

Answer 1

multiply both sides by a+4

Answer 2

when I did that I came up with 3a+16=a^2-8

Answer 3

I am currently stuck at 0=a^2-3a-24

Answer 4

3(a+4)-(a-4)=a^2-8
3a+12-a+4=a^2-8
2a+16=a^2-8
0=a^2-2a-8-16
0=a^2-2a-24 is what i get

Answer 5

may I just ask how you got the 2a, I knew that I had messed up somewhere, but I kept getting 3a and 24

Answer 6

3a-a

Answer 7

but where does that other a come from?

Answer 8

3a-a=2a
then i subtracted 2a on both sides

Answer 9

i multiplied both sides by a+4
3(a+4)-(a-4)=a^2-8

Answer 10

Thank you!! I couldnt figure out where I went wrong !!

Answer 11

how do I break it down from the point of 0=a^2-2a-24? I divided by 2 and got 12, but the final answer is supposed to be 6, could you tell me where I went wrong?

Answer 12

to factor a^2-2a-24
find two factors of -24 that have product -24
and have sum -2

Answer 13

6 and 4?

Answer 14

-6 and 4
-6(4)=-24
-6+4=-2
so we have
(x-6)(x+4)=0

Answer 15

remember we don't want a=-4 since neither function on either side is defined at x=-4
only solution is a=6

Answer 16

thank you again :]