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Mathematics 22 Online
OpenStudy (anonymous):

Show all work to simplify 2/x - 2/x-1 + 2/x-2 . Use complete sentences to explain how to simplify this expression. Remember to list all restrictions.

OpenStudy (ledah):

Crap. restrictions are going to give you some trouble

OpenStudy (anonymous):

i hate restrictions but i know how to find them from the solution :P

OpenStudy (anonymous):

get a common denominator (x)(x+1)(x+2) [2(x+1)(x+2) +2(x)(x+2) -2(x)(x+1)]/(x)(x+1)(x+2) expand and simplify [2x^2 +6x + 4 + 2x^2 +4x -2x^2 -2x]/(x)(x+1)(x+2) collect like terms [2x^2 +8x +4]/(x)(x+1)(x+2) leaves 2[x^2 +4x +2]/(x)(x+1)(x+2) does that look right ?

OpenStudy (ledah):

:O You just made me seem stupid D:< Jk

OpenStudy (phi):

looks good. now the restrictions.

OpenStudy (jiteshmeghwal9):

2/x-2/x-1+2/x-2 =2/x-2*x-1+2/x-2 =2x-2/x-1+2/x-2

OpenStudy (anonymous):

the restrictions should be ... -1 & -2 , right ?

OpenStudy (helder_edwin):

you forgot the x that is in denominator

OpenStudy (anonymous):

oh right ... umm ... shoot now im not sure ..

OpenStudy (helder_edwin):

it's fine you did x+1=0 and x+2=0 to get -1 and -2 right?

OpenStudy (anonymous):

yes

OpenStudy (helder_edwin):

ok but your denominator is \[ \large x(x+1)(x+2) \] if you put \[ \large x(x+1)(x+2)=0 \] you get x=0 or x+1=0 or x+2=0. so u missed the x

OpenStudy (anonymous):

okay thanks :)

OpenStudy (helder_edwin):

just wait!

OpenStudy (phi):

restrictions are just instructions to keep you from dividing by 0 so if you have 3/x the restriction is x≠0

OpenStudy (helder_edwin):

your original problem is \[ \large \frac{2}{x}-\frac{2}{x-1}+\frac{2}{x-2} \] correct this in what u didi before

OpenStudy (phi):

if you have 3/(x-1) you can not have x=1 because x-1 would become 1-1=0 and that is bad

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