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ronnie

New member
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Sep 2, 2005
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14
Add: (5/x) <(a fraction) + (6/x^2 -4)< (a fraction, except 4 is not part of the exponets) minus (3x/2-x)<(a fraction)


I appreciate any help, and if both problems are too much, I completely understand.


Thanks Again!
 
I'm sorry, but I don't understand your post. You have 5/x, which is less than some fraction, not given, which is added to 6/x^2, and then 4 is subtracted. This is in turn less than some other fraction, with a four subtracted somewhere, but not in an exponent somewhere. From this, you subtract 3x/2, and then subtract another x. And this is less than some other fraction which isn't named.

I have no idea what you're supposed to do with such a thing.

Eliz.
 
Stapel:

The parentheses just explain what the sets are. There is no unknown fractions, there is just those three fractions.

5overX PLUS 6overX^2-4 MINUS 3xover2-x. Period.


I appreciate your post.
 
To indicate grouping, use grouping symbols. I'm going to guess that you have the following:

. . . . .5/x + 6/(x<sup>2</sup> - 4) - 3x/2 - x

The common denominator is going to be 2x(x<sup>2</sup> - 4):

. . . . .(5/x)(2(x<sup>2</sup> - 4)/2(x<sup>2</sup> - 4)) + (6/x<sup>2</sup> - 4)(2x/2x) - (3x/2)(x(x<sup>2</sup> - 4)/x(x<sup>2</sup> - 4)) - (x/1)(2x(x<sup>2</sup> - 4)/(2x(x<sup>2</sup> - 4))

Simplify the numerators, and combine. Leave the denominator factored. Then factor the numerator, if possible, and see if anything cancels.

Eliz.
 
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