simplify
(3x+3)/(x^2+2x+1)-(x+1)/(x^2-1)
2007-01-01 14:38:12 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Algebra 2 help?
simplify
(3x+3)/(x^2+2x+1)-(x+1)/(x^2-1)
2007-01-01 14:43:01 · update #1
the last three dots are the parenthase
2007-01-01 14:43:48 · update #2
(3x+3)/(x^2+2x+1)-(x+1)/(x^2-1)
3(x+1)/(x+1)^2-(x+1)/((x+1)(x-1)
3/(x+1)-1/(x-1)
3(x-1)/(x^2-1)-(x+1)/(x^2-1)
(3x-3-x-1)/(x^2-1)
4(x-1)/((x+1)(x-1))
4/(x+1)
2007-01-01 14:43:26 · answer #1 · answered by yupchagee 7 · 15⤊ 0⤋
The common denominator is (x+1)^2(x-1)
(3x+3)/(x+1)^2(x-1)-(x+1)/
(x+1)^2(x-1)
3(x+1)-(x+1)/(x+1)^2(x-1)
3(x-1)-x-1/(x+1)^2(x-1)
3x-3-x-1/(x^2-1)
4(x-1)/(x+1)(x-1)
4/(x+1)
2007-01-01 22:52:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
(3x+3)/(x^2+2x+1)-(x+1)/(x^2-1)
= 3(x+1)/(x+1)^2 - (x+1)/[(x+1)(x-1)]
= 3/(x+1) - 1/(x-1)
=(2x-4)/(x^2-1)
2007-01-01 22:50:47 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
I belive the solution is 3/(x+1) - 1/(x-1)
2007-01-01 22:56:31 · answer #4 · answered by ? 2 · 0⤊ 0⤋
factor everything. you get
3(x+1)/(x+1)² - (x+1)/[(x+1)(x-1)]
then cancel common factors in the first fraction, getting
3/(x+1) - (x+1)/[(x+1)(x-1)] =
3(x-1)/(x²-1) - (x+1)/(x²-1) =
(3x - 3 - x - 1)/(x²-1) =
(4x + 2)/(x²-1)
2007-01-01 22:52:40 · answer #5 · answered by Philo 7 · 0⤊ 0⤋
3(x+1)/(x+1)^2 - (x+1)/(x+1)(x-1)
3(x-1)/(x+1)(x-1) -(x+1)/(x+1)(x-1)
3x-3/(x+1)(x-1) -x+1/(x+1)(x-1)
3x-1 -(x-1)/(x+1)(x-1)
2x/(x+1)(x-1)
2007-01-01 23:11:16 · answer #6 · answered by chevygirl0130 1 · 0⤊ 0⤋
It would help if u typed the whole problem out then we could see everything that we're dealing with.
But try starting with the order of operations and then take it from there.
P.S. If u don't know what the order of operations are then u probably shouldn't be in Algebra II.
2007-01-01 22:45:29 · answer #7 · answered by Hmmm Andrew ♣ 2 · 0⤊ 0⤋
You have to type to whole problem out.
2007-01-01 22:40:55 · answer #8 · answered by Dalonna 2 · 1⤊ 0⤋