Thursday
April 24, 2014

Homework Help: Rational Expression

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

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

TO KIM - can you help me with this |2x - 3|^2 - 6|2x - 3| + 5=0 I know you are ...
Math33333 - Please help me! Problem: x+2 x-3 _________ - _________ 2x^2+5x+2 2x^...
Algebra - Need help with this problem, 2x+y=6 solve for y 2x+y=6 subtract 2x ...
Algebra help - Find (fog)(x)and (gof)(x)if f(x)=x-x^2 and g(x)= 2x+3. Okay so I ...
Algebra - I multiplyed x^2-2x+2 times x^2+2x+2. The book suggested the order ...
check my work on derivatives - Can someone check my work? So, I took the ...
Math - What is the first step. Explain please. Which expression is equivalent to...
Math - Can someone please check this for mistakes please? y=4(x+3)-2x y=4x+3-2x ...
Math - Prove each identity: a) 1-cos^2x=tan^2xcos^2x b) cos^2x + 2sin^2x-1 = sin...
calculus - So I am suppose to evaulate this problem y=tan^4(2x) and I am ...

Search
Members