The problem is given as

We do not know if 6x–1 is non-negative or negative. Therefore we divide the problem into two cases, one where 6x–1 is non-negative and one where 6x–1 is negative.

Case 1: 6x–1 0

Case 2: 6x–1 0

In this case we know that 6x–1 0. First we note that this means that 6x 1, or x 1/6. Second, by the definition of absolute value, we know that for this case |6x – 1| can be replaced by 6x – 1. Now our problem appears as  6x – 1 = |2x – 3| We still have an absolute value here. We do not know if 2x – 3 is non-negative or negative. Therefore, we divide the problem again. Case 1.1: 2x–3 0 Case 1.2: 2x–3 0 In this case we know that 2x–3 0. This means that 2x 3, or x3/2. This is consistent with the fact that we know that x1/6 in order to be in Case 1. In addition, because 2x–3 0, we know from the definition of absolute value that |2x–3| can be replaced by 2x–3. Now our problem becomes 6x – 1 = 2x – 3 which becomes 4x = – 2 or x = – 1/2 Unfortunately, this value is inconsistent with the Case 1 restriction that x1/6. Therefore, there is no solution in this case. In this case we know that 2x–3 0. This means that 2x 3, or x3/2. This is consistent with the fact that we know that x1/6 in order to be in Case 1. In addition, because 2x–3 0, we know from the definition of absolute value that |2x–3| can be replaced by – (2x–3). Now our problem becomes  6x – 1 = – (2x – 3) which becomes 6x – 1 = – 2x + 3 and that becomes 8x= 4 or x=1/2 This is consistent with both the Case 1 condition that x1/6 and the Case 1.2 condition that x3/2. Therefore, x=1/2 is a soltuion.

In this case we know that 6x–1 0. First we note that this means that 6x 1, or x 1/6. Second, by the definition of absolute value, we know that for this case |6x – 1| can be replaced by – (6x – 1.) Now our problem appears as – (6x – 1) = |2x – 3| We still have an absolute value here. We do not know if 2x – 3 is non-negative or negative. Therefore, we divide the problem again. Case 2.1: 2x–3 0 Case 2.2: 2x–3 0 In this case we know that 2x–3 0. This means that 2x 3, or x3/2. This is inconsistent with the fact that we know that x1/6 in order to be in Case 2. Therefore, there is no possible solution for this case. We can not have an answer that is both 1/6 and 3/2. In this case we know that 2x–3 0. This means that 2x 3, or x3/2. This is consistent with the fact that we know that x1/6 in order to be in Case 2. In addition, because 2x–3 0, we know from the definition of absolute value that |2x–3| can be replaced by – (2x–3). Now our problem becomes – (6x – 1) = – (2x – 3) This is the same as 6x – 1 = 2x – 3 which gives 4x = – 2 or x = – 1/2 This answer is consistent with the Case 2 condition that x1/6 and with the Case 2.2 condition that x3/2. Therefore, x=– 1/2 is a solution.

The table above produces the answers x=1/2 and x=^– 1/2. The table represents a general soltuion to such problems in that it divides the problem into numerous cases such that we can replace the absolute value expressions with non-absolute value expressions. As a small note, the length and complexity of the table comes from the extensive commentary placed into it.