Question

simplify the sum.
7/a+8 + 7/a^2-64

Answer 1

\[\frac{ 7 }{ a+8 } + \frac{ 7 }{ a^{2}-64 }\]

Answer 2

First factor the second denominator
\[\Large \frac{7}{a+8} + \frac{7}{a^2-64}\]
\[\Large \frac{7}{a+8} + \frac{7}{(a-8)(a+8)}\]
What's the next step?

Answer 3

im not sure. can you walk me through it?

Answer 4

if you wanted to add fractions like 1/2 and 2/3, how would you do it?

Answer 5

oh i know. (a-8)(a+8) would be a^2-64

Answer 6

that is correct

Answer 7

do you see why I factored?

Answer 8

so now its\[\frac{ 7 }{ a^{2}-64 }\]

Answer 9

yes i see. whats the next step,?

Answer 10

how do you add fractions? what is the key thing needed?

Answer 11

umm not sure

Answer 12

something to do with the denominators

Answer 13

idk :/

Answer 14

have a look at this page

http://www.mathsisfun.com/fractions_addition.html

Answer 15

hopefully that's all review

Answer 16

so ive gotta make the denominators equal?

Answer 17

exactly

Answer 18

the denominators at this point are (a+8) and (a-8)(a+8)

what's missing from the first denominator?

Answer 19

i though the denominators where (a+8) and (a^2-64) now...

Answer 20

that's what they were originally

Answer 21

but I factored a^2-64

Answer 22

oh. right. sorry i was confused

Answer 23

the denominators at this point are (a+8) and (a-8)(a+8)

what's missing from the first denominator?

Answer 24

(a-8) ?

Answer 25

so you have to multiply top and bottom of the first fraction by (a-8) to get that first denominator equal to the second

Answer 26

\[\Large \frac{7}{a+8} + \frac{7}{a^2-64}\]
\[\Large \frac{7}{a+8} + \frac{7}{(a-8)(a+8)}\]
\[\Large \frac{7(a-8)}{(a-8)(a+8)} + \frac{7}{(a-8)(a+8)}\]

Answer 27

What's next?

Answer 28

hmmm. does anything cancel out?

Answer 29

i honestly have no idea. i cant figure it out

Answer 30

@jim_thompson5910

Answer 31

are the denominators the same at this point?

Answer 32

yes

Answer 33

so you can add up the numerators and place that over the common denominator

\[\Large \frac{7(a-8)}{(a-8)(a+8)} + \frac{7}{(a-8)(a+8)}\]
\[\Large \frac{7(a-8)+7}{(a-8)(a+8)}\]

I'll let you simplify from here

Answer 34

\[\frac{ 7a-49 }{ (a-8)(a+8) }\]

Answer 35

you nailed it

Answer 36

optionally you can expand out the denominator to get

\[\Large \frac{7a-49}{(a-8)(a+8)}\]

\[\Large \frac{7a-49}{a^2 - 64}\]