Posted by Mandi on Monday, October 16, 2006 at 9:56pm.
(x+1/2x-1 - x-1/2x+1) * (2x-1/x - 2x-1/x^2)
(x+1/2x-1 - x-1/2x+1)*(2x-1/x - 2x-1/x^2)
The first parenthesis reduces to zero. Check that.
I have the answers in the back of my book (just not how to get there) the book shows the answer as
6(x-1)/x(2x+1). When I solve, I get
6(2x-1)/(2x+1), but I kinda stink at this stuff.
OK, I think I understand what you meant to type:
[(x+1)/(2x-1) - (x-1)/(2x+1) ]* [(2x-1/x - (2x-1)/x^2)]
common denominator in first [] is (2x-1)(2x+1)
first bracket only.
6x/(2x+1)(2x-1)*[(2x-1)(x-1) ]/x^2 [(x+1)(2x+1) - (x-1)*(2x-1) ]/(2x-1)(2x+1)
[2x^2+3x+1 - 2x^2+3x -1]/(2x+1)(2x-1)
6x/(2x+1)(2x-1)
second bracket:
[(2x-1/x - (2x-1)/x^2)]
[(2x^2-x - 2x+1 ]/x^2
[(2x^2-3x +1 ]/x^2
[(2x-1)(x-1) ]/x^2
combining the two brackets...
6x/(2x+1)(2x-1)*[(2x-1)(x-1) ]/x^2
The 2x-1 divides out..
6x/(2x+1) *[ (x-1) ]/x^2 as does x
6 /(2x+1) *[ (x-1) ]/x
6(x-1)/x(2x+1)
Ok, Thank you so much for your help!
1-2/x divided by x+4/9x
How do I solve this problem?
What is Rational Expressions?
5 8
____ + ____
y-3 3-y
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