Question

Add.simplify if possible.
4w/w^2-64 + w/w-8 =

Answer 1

Hi Alisa, lets have a look.

Answer 2

The first thing you could realize, is that you have a special form of a polynomial on the bottom of the first term, it is in the form: \[a^2-b^2=(a+b)(a-b)\]
Or in your case, it is \[(w)^2-(8)^2 = (w-8)(w+8)\]

Answer 3

Next, do not worry about common denominator as everyone does, all you have to do is use this formula for adding fractions:
\[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]

Answer 4

In your case a is 4w, b is (w-8)(w+8), c is w, and d is (w-8), so you get:
\[\frac{4w}{w^2-64}+\frac{w}{w-8}=\frac{4w}{(w+8)(w-8)}+\frac{w}{w-8}==\frac{4w(w-8)+w(w-8)(w+8)}{(w-8)(w+8)(w-8)}\]

Answer 5

factor out (w-8) in the numerator, and cancel terms.

Answer 6

See I got the 4w/(w+8)(w-8)

Answer 7

I'm confused...

Answer 8

Hey, sorry about that I had to run off for a bit.

Answer 9

I was like hey you left me hanging lol

Answer 10

I came up with 4w/(w+8)(w-8)

Answer 11

Let me check real quick, also you can check if you are correct by plugging a number in for w (say 1 or something) into your simplified form and the original. They should be the same if you did it correctly

Answer 12

Lets try that, if you plug in 1 for your answer we get -(4/63)

Answer 13

If we plug in 1 to the original we get something different (I just typed it into my calculator)

Answer 14

So you made an error somewhere. why dont you show me your steps and I can help you along the way.

Answer 15

Ok, well I can try to explain it again. So you figured out that you can factor the w^2-64, this will leave you with \[\frac{4w}{(w-8)(w+8)}+\frac{w}{(w-8)}\]

Answer 16

ok so you put 4w(w-8)+w(w-8) / (w-8)(w+)(w-8)
so I got 4w/(w+8)(w-8)... I just added

Answer 17

Or is it 4w^2-32/w-8

Answer 18

Look at what I just put, this is the simple formula you can use for adding fractions.
\[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]

Answer 19

Ok thanks!!

Answer 20

In this is case, "a" is equal to 4w right?

Answer 21

"b" is equal to (w+8)(w-8) right?

Answer 22

"c" is equal to w, and "d" is equal to (w-8)

Answer 23

So plugging that into the formula you get
\[\frac{4w(w-8) + w(w+8)(w-8)}{(w+8)(w-8)(w-8)}\]

Answer 24

Now you can factor (w-8) out of the numerator, since it appears in both terms, this gives you:\[\frac{(w-8)[4w+w(w+8)]}{(w+8)(w-8)(w-8)}\]

Answer 25

Cancel (w-8) from the numerator and denominator

Answer 26

\[\frac{42 + w(w+8)}{(w+8)(w-8)}\]

Answer 27

42 should be 4w*

Answer 28

Then expand what you have left, since there is no more factoring or simplifying that can be done.

Answer 29

\[\frac{w^2+12w}{w^2-64}\]

Answer 30

That is as simplified as it can get

Answer 31

Ok thanks alot! I'll use your method on similiar problems I have

Answer 32

Cool, yeah I find that it is a lot easier than worrying about the stupid least common denominator etc. It also works on just regular fractions with numbers.