Mathematics
8 Online
OpenStudy (anonymous):
Please help me understand this.
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OpenStudy (anonymous):
There is nothing to understand?
OpenStudy (anonymous):
\[\frac{2}{x-1}+1 = \frac{2}{x ^{2}-x}\]
OpenStudy (anonymous):
Solve for x:
OpenStudy (anonymous):
I did this whole problem for you an hour ago. You didn't understand it?
OpenStudy (anonymous):
I know you tried to help me earlier Mathandphysics, but I didn't understand how you got the common denominator
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OpenStudy (anonymous):
the common denominator is something that can be shared by all 3 denominators. So what does (x-1), x, and (x^2 - x) have in common?
OpenStudy (anonymous):
I tried to pull it from the left side and I keep getting (x-1)(x^2-x) as the common denominator
OpenStudy (anonymous):
they are all part of (x^2 - x) because x times (x - 1) will get you to that.
OpenStudy (anonymous):
(x-1) is already part of (x^2 - x) so you don't need it
OpenStudy (anonymous):
We could cheat...
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OpenStudy (anonymous):
So you can just throw it out the window?
OpenStudy (anonymous):
you don't throw it out. it is INCLUDED INSIDE of (x^2 - x)
OpenStudy (anonymous):
Let's cheat
OpenStudy (anonymous):
Let's do the left side only
OpenStudy (anonymous):
ok which would be 2/x-1 + x-1/x-1 right?
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OpenStudy (anonymous):
Yes, it's all over x-1
OpenStudy (anonymous):
but wouldn't leave you with just 2/x-1
OpenStudy (amistre64):
\[\frac{2(x)}{x-1}+1(x) = \frac{2(x)}{x ^{2}-x}\]
\[\frac{2x}{x-1}+x = \frac{2x}{x-1}\]
OpenStudy (anonymous):
sat i think ur on the wrong question man :)
OpenStudy (anonymous):
Complicated it all again...
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OpenStudy (anonymous):
fakshon look ..
OpenStudy (anonymous):
if you had 1/2 + 1/4 = 1/6 ... what would be the common denominator?
OpenStudy (amistre64):
i forgot to cancel my x on the right :) doh!!
OpenStudy (anonymous):
ok which would be 2/x-1 + x-1/x-1 right?
ie 1+x/x-1
OpenStudy (anonymous):
Just follow along, OK...
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OpenStudy (anonymous):
oh my page was messed up .. thats weird
OpenStudy (anonymous):
So now we have
1+x/x-1 = 2/x^2-x agreed?
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
Do u know how to cross multiply?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
Go on then, do it
OpenStudy (anonymous):
can you cross multiply over an = sign? Never done that before.
OpenStudy (anonymous):
you HAVE to cross multiply across an equal sign
OpenStudy (anonymous):
ie a/b =c/d -> ad = bc
OpenStudy (anonymous):
I'll do it
(1+x)(x^2-x) = 2(x-1)
U follow?
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OpenStudy (anonymous):
Yes? No?
OpenStudy (anonymous):
ok so x^3-1 = 2x-2
OpenStudy (anonymous):
U went to fast....
OpenStudy (anonymous):
x^2-x is equal to x(x-1), right?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
So you can cancel the left and right of x-1
OpenStudy (anonymous):
Leaving
(1+x)x = 2
OpenStudy (anonymous):
so x = 2 not -2?
OpenStudy (anonymous):
or did I lose it, lol
OpenStudy (anonymous):
Now we expand
(1+x)x = 2 ->
x+x^2 = 2 which is a quadratic...
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OpenStudy (anonymous):
hey estudier go to chat after ur done
OpenStudy (anonymous):
Quadratic factors as (x+2)(x-1) _> x= -2 or 1