I hope this helps
http://img485.imageshack.us/img485/9931/fazde7.jpg
2007-03-16 16:42:00
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answer #1
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answered by M. Abuhelwa 5
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All you have to do is turn the problem into a multiplication problem. Instead, do (x+2)/x x 3x/(x-2) which is the same thing as what you said.
From that problem, you can cancel the x from 3x in the numerator and the x in the denominator.
That leaves you with the problem (x+2)/1 x 3/(x-2).
So you get 3(x+2)/(x-2)
Hope that helps!
2007-03-16 23:31:51
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answer #2
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answered by jkt610 2
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So it looks like you are dividing fractions, right?? Just remember, "when dividing fractions, don't be shy. Just flip over the second number and multiply." It's so simple as multiplying by the reciprocal. So,
((x+2)/ x) (3x/ (x-2))
You can use cross products to eliminate an x in the denominator of the first fraction and the numerator of the second fraction. So,
((x+2)/ 1) (3/ (x-2)
Then multiply the numerators together and the denominators together, So,
(3(x+2)) / (x-2) or (3x+6) / (x-2)
2007-03-16 23:25:34
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answer #3
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answered by Cameron K 2
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= [(x+2)/x] / [(x-2)/3x]
Get the reciprocal of [(x-2)/3x] and multiply it to [(x+2)/x]
= [(x+2)/x] * [3x/(x-2)]
you can cancel x and 3x so
= (x+2)* [3/(x-2)]
multiply 3 to (x+2)
= (3x+6)/(x-2) is the answer
2007-03-16 23:29:24
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answer #4
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answered by PANCAKE LOVER 04 1
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make it a (x) problem
(x+2)/x TIMES (3x)/(x-2)
x cancels out so i guess yr left with...
(3x+6)/(x-2)
2007-03-16 23:22:29
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answer #5
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answered by .: ZEIDO :. 3
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