Question

can
(25h-8)/((3h)(h-2))
be simplified?

Answer 1

doesn't look like it to me...

Answer 2

\[\frac{ 25h-8 }{ 3h ^{2}-6h }\]

Answer 3

then i factored out a
3h..

Answer 4

yeah I don't see much more you can do with it.

Answer 5

me either

Answer 6

so i got (25h-8)/3h(h-2)

Answer 7

i think that's as good as it gets

Answer 8

but that' s still not the right answer.... my program isn't taking it

Answer 9

\[\frac{ 25 }{3h-6 } - \frac{ 8 }{ 3h ^{2} -6h}\]

Answer 10

this is an expression though so you can't multiply by 3h

Answer 11

How did you get that? and You can multiply by 3h/3h which is 1

Answer 12

but then you'd be back at the same expression

Answer 13

actually I see what your saying you then want to take out 1/3h right

Answer 14

the original problem is
(21h+4h-8)/((3h)(h-2))

Answer 15

why didnt you post that from the beginning :/

Answer 16

Ya, I deleted my post because I think it was a dead end

Answer 17

so that gives you
(25h-8)/((3h)(h-2))

Answer 18

oh I think I see an answer

Answer 19

There must be a reason 21h and 4h is split up.

Answer 20

you need to get h-2 through factorization

Answer 21

\[\frac{ 21h +4(h-2) }{3h(h-2) }\]

Answer 22

\[\frac{ 21h + (h-2)(4) }{ 3h(h-2) }\]

Answer 23

haha

Answer 24

split up into 2 different fractions next

Answer 25

that one got me

Answer 26

now try to solve it mberdeja

Answer 27

so it's
(21+4)/(3h)

Answer 28

(21h+4)/(3h)

Answer 29

no you have to break up the fraction.

Answer 30

\[\frac{ 21 }{ 3h-6 } + \frac{ 4 }{ 3h }\]

Answer 31

no cause the h-2 cancels at the bottom and at the top

Answer 32

your right but if you look the top is conjoined by an addition operator

Answer 33

you cant just cancel terms unless the term is multiplied by the entire thing which it isn't