Rational Expression
posted by Mandi .
(x+1/2x1  x1/2x+1) * (2x1/x  2x1/x^2)
(x+1/2x1  x1/2x+1)*(2x1/x  2x1/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(x1)/x(2x+1). When I solve, I get
6(2x1)/(2x+1), but I kinda stink at this stuff.
OK, I think I understand what you meant to type:
[(x+1)/(2x1)  (x1)/(2x+1) ]* [(2x1/x  (2x1)/x^2)]
common denominator in first [] is (2x1)(2x+1)
first bracket only.
6x/(2x+1)(2x1)*[(2x1)(x1) ]/x^2 [(x+1)(2x+1)  (x1)*(2x1) ]/(2x1)(2x+1)
[2x^2+3x+1  2x^2+3x 1]/(2x+1)(2x1)
6x/(2x+1)(2x1)
second bracket:
[(2x1/x  (2x1)/x^2)]
[(2x^2x  2x+1 ]/x^2
[(2x^23x +1 ]/x^2
[(2x1)(x1) ]/x^2
combining the two brackets...
6x/(2x+1)(2x1)*[(2x1)(x1) ]/x^2
The 2x1 divides out..
6x/(2x+1) *[ (x1) ]/x^2 as does x
6 /(2x+1) *[ (x1) ]/x
6(x1)/x(2x+1)
Ok, Thank you so much for your help!
12/x divided by x+4/9x
How do I solve this problem?
What is Rational Expressions?
5 8
____ + ____
y3 3y
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