Question

8x/x+1 - 3/x+1

Answer 1

What are you trying to figure out?

Answer 2

@Hero @ParthKohli @RANE

Answer 3

I'm just adding and subtracting.

Answer 4

Are you trying to solve for x or something?

Answer 5

@hutchkat

Answer 6

Number 17

Answer 7

If you zoom in on it you can see it fine...

Answer 8

when denominators are common, you can just combine the numerators
like
\(\huge \dfrac{☺}{♥}+\dfrac{☻}{♥}=\dfrac{☺+☻}{♥}\)

Answer 9

but there is a variable...??

Answer 10

\(\huge \dfrac{8x}{♥}-\dfrac{3}{♥}=\dfrac{8x-3}{♥}\)
whatever be the denominators....it'll be same.

Answer 11

for your problem,
\(\huge \dfrac{8x}{x+1}-\dfrac{3}{x+1}=\dfrac{8x-3}{x+1}\)
thats it!

Answer 12

ok thnks @hartnn and @hutchkat . Now what about 19

Answer 13

Number 19

Answer 14

Scroll up to see my attachment.

Answer 15

http://assets.openstudy.com/updates/attachments/51af3b89e4b05b167ed23c22-pyromancer-1370438863548-here.jpg

Answer 16

You need to make the denominator like terms. So what do you need to multiply 2+x and/or 4-x by to make them the same? And remember, whatever you do to part of an equation you must do to the whole equation.

Answer 17

wow hatnn is in love xD

Answer 18

bdw, it wasn't 5 for 17th, just (8x-3)/(x+1)
if denominators are not same, you can cross multiply..
\(\huge \dfrac{☺}{♥}+\dfrac{☻}{♦}=\dfrac{☺♦+☻♥}{♥}\)

Answer 19

lol, no...just so a pictorial view of fraction addition/subtraction...

Answer 20

ok

Answer 21

do this,
\(\huge \dfrac{☺}{♥}+\dfrac{☻}{♦}=\dfrac{☺♦+☻♥}{♥♦ }\)

Answer 22

then simplify