3x + 4y + 2y = 1
The solutions for x, y, and z is { ( (1/3-4l-2m) | 3l | 2m ) }, where y = l and z = m.
I've tried this method, presupposing y = l and m = z, then it came to
I. l = (1 - 3x -2m) / 4
II. m = (1 - 3x -4l) / 2
If I try to put in either (I) or (II) to x, it would come...
Log x ((x+3)/(x-1) > Log x x ??
I've managed to find 4 conditions for this inequality:
1. -1 > x > 3
2. x > -3
3. x > 0
4. x ≠ 1
but I'm not sure how to write the solution. Is it " 0 < x & 1 < 0 < 3 " ?
Thanks.