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Mathematics 20 Online
OpenStudy (anonymous):

How will we solve this?? (3/x)+ (3/(x+1) - (3)/(x+2) please explain how to get the new denominator and any restrictions..thanks

OpenStudy (anonymous):

you need the same factors in the denominator... what would you do if you had 1/3+ 1/5?

OpenStudy (anonymous):

multiply 1/3 by 5 and 1/5 by 3

OpenStudy (anonymous):

exactly.. just do the same

OpenStudy (anonymous):

so the denominator would have x(x+1)(x+2)

OpenStudy (lalaly):

\[(\frac{3}{x} \times \frac{(x+1)(x+2)}{(x+1)(x+2)})+(\frac{3}{x+1} \times \frac{x(x+2)}{x(x+2)})-(\frac{3}{x+2}\times \frac{x(x+1)}{x(x+1)})\]

OpenStudy (anonymous):

*trying to absorb this*

OpenStudy (anonymous):

oooh i see whats going on...

OpenStudy (anonymous):

can we simplify this to this --> 3(x^2 +4x+2)/ x(x+1)(x+2)

OpenStudy (anonymous):

dont let the binomials scare you.. they are just factors... oh and unless there is a equal sign suggesting an equality you are not solving for any unknown

OpenStudy (anonymous):

there is no equal sign? does that mean i have to solve for the unknown?

OpenStudy (anonymous):

i wouldnt think so.. you are just following the instructions given for the problem.. ie simplify or whatever they want you to do

OpenStudy (anonymous):

it says mention if there are any restrictions?

OpenStudy (anonymous):

from what i remember..restrictions were where you had to plug in 0 in the denominator?!?

OpenStudy (lalaly):

the denominator should not be equal to zero

OpenStudy (lalaly):

\[x \cancel{=} 0,-1,-2\]

OpenStudy (anonymous):

right

OpenStudy (anonymous):

ooh..ok...

OpenStudy (anonymous):

i got it...

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