Alternate solution to Chapter 1, Section 2, Example 22

The problem is given as

|6x – 1| = |2x – 3| The solution in the text squares both sides of the equation to eliminate the absolute value. This page solves the problem using the definition presented in the absolute value web page, namely,

We use this definition to change absolute value problems into problems that do not have an absolute value.

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.