je suppose que les fractions inclus plusieurs termes :
(2x+1)/9 + (1-x)/6 = (3x-2)/18 - x/3
(2x+1)/9 + (1-x)/6 - (3x-2)/18 + x/3 = 0
2(2x+1)/18 + 3(1-x)/18 - (3x-2)/18 + 6x/18 = 0
2(2x+1) + 3(1-x) - (3x-2) + 6x = 0
4x + 2 + 3 - 3x - 3x+ 6 + 6x = 0
6x + 6x - 3x - 3x + 6 + 3 + 2 = 0
6x + 11 = 0
6x = -11
x = -11/6
En espérant t'avoir aidé.
