(2X / X+3) + (3 / X-3) = 8 / X^2 -9
2007-11-21 08:05:34 · 5 answers · asked by D*Star 1 in Science & Mathematics ➔ Mathematics
2x / (X+3) + 3 / (x-3) = 8 / (x^2 -9)
[2x(x-3) + 3(x+3) - 8] / (x+3)(x-3) = 0
(2x² - 6x +3x + 9 -8] / (x+3)(x-3) = 0
(2x² -3x +1) / (x+3)(x-3) = 0
Then x diferent of 3 and -3 and 2x² -3x +1 = 0
D = 9 -4*1*2 = 1
x = (3 +/-1)/4 : x = 1 or x = 1/2
2007-11-21 08:15:32 · answer #1 · answered by antone_fo 4 · 0⤊ 0⤋
(2x / x+3) + (3 / x-3) = 8 / x^2-9
(2x / x+3) + (3 / x-3) = 8 / (x-3)(x+3)
So, multiply everything by the common denominator (x - 3)(x + 3)
(x - 3)(2x) + (x + 3)(3) = 8
2x^2 - 6x + 3x + 9 = 8
2x^2 - 3x + 1 = 0
(2x - 1)(x - 1) = 0
x = 1 and 1/2
2007-11-21 08:12:30 · answer #2 · answered by Mathematica 7 · 0⤊ 0⤋
2x/(x + 3) + 3/(x -- 3) = 8/(x^2 -- 9)
=> 2x(x -- 3) + 3(x + 3) = 8
=> 2x^2 -- 3x + 1 = 0
=> 2x^2 -- 2x -- x + 1 = 0
=> 2x(x -- 1) -- 1(x -- 1) = 0
=> (x-- 1)(2x -- 1) = 0
giving x = 1, 1/2 answer.
2007-11-21 08:18:40 · answer #3 · answered by sv 7 · 0⤊ 0⤋
2x/(x+3) + 3/(x-3) = 8/(x^2-9)
[2x(x-3) + 3(x+3)]/(x+3)(x-3) = 8/(x^2-9)
[2x^2 - 6x + 3x + 9]/(x^2-9) = 8/(x^2-9)
[2x^2 - 3x + 9] = 8
2x^2 - 3x + 1 = 0
2x^2 -2x - x + 1 = 0
2x(x - 1) - 1(x - 1) = 0
(x - 1)(2x - 1) = 0
x = 1 or 1/2
2007-11-21 08:14:58 · answer #4 · answered by mohanrao d 7 · 0⤊ 0⤋
2x/x+3 + 3/ x-3=8 / x^2-9
2x(x-3) +3(x+3)=8
2x^2-6x+3x+9=8
2x^2-3x+1=0
(2x-1)(x-1)=0
x=1/2; 1
2007-11-21 08:21:54 · answer #5 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋