Posted by
**Mandi** on
.

(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