View Full Version : Simplifying Negatives part 2
hahajamiexlyn
Aug 22, 2007, 06:58 AM
Okay I figured out all of the problems in my packet except
-4(2x-1/2x+1)^-3[2(2x+1)-2(2x-1)/(2x+1)^2]
:eek:
ebaines
Aug 22, 2007, 07:14 AM
Please clarify - is this the expression you're trying to simplify?
-4 \left( \frac {2x-1} {2x+1} \right) ^ {-3} \left{ 2(2x+1) - 2 \frac {(2x-1)} {(2x+1)^2} \right}
hahajamiexlyn
Aug 23, 2007, 06:30 AM
Please clarify - is this the expression you're trying to simplify?
-4 \left( \frac {2x-1} {2x+1} \right) ^ {-3} \left{ 2(2x+1) - 2 \frac {(2x-1)} {(2x+1)^2} \right}
no it's
http://i14.tinypic.com/6byd541.jpg
ebaines
Aug 23, 2007, 02:29 PM
In your original post you showed the first term as (2x-1/2x+1)^-3, but in the repost you have it as (2x+1)^3/(2x-1) -- please confirm that the repost is correct. Given that the subject title is about negative exponents, I'm wondering if a typo or two didn't creep into the repost?
Assuming the repost is correct, I would suggest first combining the stuff in the numerator inside the brackets: 2(2X+1)-2(2x-1). Multiply through and this becomes simply 4. Then look at the (2x+1) terms in the denominators - you have (2X+1)(2x+1)^2 -- this is the same as (2x+1)^3. Do you see that? Can you take it from there?