Math Problem - 2

jeffeh

Banned
Dec 8, 2004
304
0
0
( 2x-2 / x^2 )* (x^3 / 4x - 8)

so from here I expanded and cancelled out (x-1) in the first fraction

then combined the two to get

2x^3 / 4x^2-2

from here I'm not sure if the x^3 would cancel out the x^2 on the bottom?
 

kevinthenerd

Platinum Member
Jun 27, 2002
2,908
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Originally posted by: Syringer
Originally posted by: ntdz
Just cross multiply, it's easier to do it that way...

Are you being serious? You need an equal sign to cross multiply.

Maybe they're trying to mislead him? I was honestly tempted to do so through a PM. lol
 

ColdFusion718

Diamond Member
Mar 4, 2000
3,496
9
81
Originally posted by: kevinthenerd
Originally posted by: Syringer
Originally posted by: ntdz
Just cross multiply, it's easier to do it that way...

Are you being serious? You need an equal sign to cross multiply.

Maybe they're trying to mislead him? I was honestly tempted to do so through a PM. lol

How would you feel if someone purposely mislead you when you asked for help in one of your EE classes?
 

jagec

Lifer
Apr 30, 2004
24,442
6
81
Would it kill you to use parenthesis? (x^3 / 4x - 8) != (x^3 / (4x - 8))

anyway, you might want to redo your work...there's only one (x-1) term, so you can't cancel.