2007-06-28 11:17:07 · 4 answers · asked by Russel 1 in Science & Mathematics ➔ Mathematics
Multiply both sides by (2x+1), then by (x-3). This gets rid of the fractions. Then expand, simplify, and solve.
(2x+1)/(2x+1) = (2x+1)/(x-3) + 2(2x+1)
1 = (2x+1)/(x-3) + 2(2x+1)
(x-3) = (2x+1) + 2(2x+1)(x-3)
-x - 4 = 2(2x^2 - 5x - 3)
-x - 4 = 4x^2 - 10x - 6
0 = 4x^2 - 9x - 2
etc.
2007-06-28 11:19:56 · answer #1 · answered by Anonymous · 0⤊ 0⤋
(2x+1)-¹ = (x-3)-¹ + 2
1 over 2x + 1 = 1 over x-3 + 2
2007-06-28 11:21:26 · answer #2 · answered by aeiou 7 · 0⤊ 1⤋
(2x+1)^-1=(x-3)^-1+2
Or 1/(2x+1)=1/(x-3)+2
Cross Multiplying
(x-3)=(2x+1)+2(2x+1)(x-3)
Solve the quadratic for x
2007-06-28 11:32:31 · answer #3 · answered by MAHAANIM07 4 · 0⤊ 0⤋
1 / (2x + 1) - 1 / (x - 3) = 2
(x - 3 - 2x - 1) / (2x + 1).(x - 3) = 2
(- x - 4) = 2.(2x² - 5x - 3)
4x² - 9x - 2 = 0
x = [ 9 ± â113 ] / 2
x = 9.8 , x = - 0.79
2007-07-02 06:39:07 · answer #4 · answered by Como 7 · 0⤊ 0⤋