3x+5/5x-3 - 5x-3/3x+5 over 5x-3/3x+5 - 3x+5/5x-3
If you can solve that, that would be great lol.
2006-10-18 16:30:45 · 5 answers · asked by severedangel1776 1 in Science & Mathematics ➔ Mathematics
-1 would be the answer. This problem is MUCH easier than it seems.
Replace 3x+5 by u and 5x-3 by v
u/v - v/u
----------
v/u - u/v
then make it even easier:
u/v = 2
v/u = 1/2
2 - 1/2
---------
1/2 - 2
= -1 :)
2006-10-18 16:33:16 · answer #1 · answered by icez 4 · 0⤊ 0⤋
[(3x+5)/(5x-3) - (5x-3)/(3x+5)]/[(5x-3)/(3x+5) - (3x+5)/(5x-3)]
Here we have a big fraction, the big numerator and the big denominator each being the difference of 2wo small fractions.
The LCD of all the small fractions is (5x - 3)(3x + 5)
Multiply the big numerator and big denominator by this LCD to clear the small fractions
Wait a sec - this is what you might do if the problem had a different complexion. Here, however, if you factor out a -1 from the big numerator you can write
- [- (3x+5)/(5x-3) + (5x-3)/(3x+5)]/[(5x-3)/(3x+5) - (3x+5)/(5x-3)]
= -1
2006-10-18 23:46:15 · answer #2 · answered by kindricko 7 · 0⤊ 0⤋
Too simple
2006-10-18 23:35:41 · answer #3 · answered by deepak57 7 · 0⤊ 0⤋
-2x + 2 over 8x - 2 is as far as i can remember how to do it..
2006-10-18 23:36:41 · answer #4 · answered by me_rini89 1 · 0⤊ 0⤋
i dont have that kind of time
2006-10-18 23:34:44 · answer #5 · answered by cmac4224 2 · 0⤊ 0⤋