Follow the same procedure I outlined in your last question.
using MarkFL2's clever approach, note that |a| = √(a^{2}).
so |1-x| < |2x-5| is the same as:
√(1-x)^{2} < √(2x-5)^{2}
squaring both sides:
(1-x)^{2} < (2x-5)^{2}
1-2x+x^{2} < 4x^{2}-20x+25
0 < 3x^{2}-18x+24
0 < x^{2}-6x+8
0 < (x-2)(x-4) <---for this to be true, both factors must be either both positive, or both negative.