Note that the denominator cannot be zero, so it has to be positive as a modulus. As such, we can simply cross multiply the term.

(For your info, if the the signage of the denominator is not known, or if it can be positive or negative, we cannot simply cross multiply them, but this is higher level stuff)

3x - 1 >= 4 |x - 1|
4 |x - 1| <= 3x - 1
|4x - 4| <= 3x - 1

An easy idea from here is to solve this inequality as a regular = without the < followed by plotting the graph.

Of course, because of the constraint (in every modulus expression) that the output is always non-negative, the minimal condition is 3x - 1 = 0 or x >= 1/3.

However, x cannot equal 1 since the original expression had |x - 1| as its denominator.

Now, at the point where the graph y = |4x - 4| intersects the line y = 3x - 1,

|4x - 4| = 3x - 1
4x - 4 = 3x - 1 or 4x - 4 = 1 - 3x
x = 3 or 7x = 5
x = 3 or x = 5/7

Let’s split these into the regions x < 5/7, 5/7 < x < 3 and x > 3.

If you were to plot the graphs together,
- for the region x < 5/7, the modulus graph is above the line
- for the middle region 5/7 < x < 3, the line is above the modulus graph
- for the region x > 3, the modulus graph is above the line

Our inequality is such that 3x - 1 >= |4x - 4|, so we want the region(s) where the line is above the modulus graph.

Hence, we have

5/7 <= x <= 3 except for x = 1

This condition is already above the mini condition of x >= 1/3 which we have established earlier.

ie 5/7 <= x < 1, 1 < x <= 3

thank you so much for such a detailed explanation !!