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## Homework Statement

Solve the equation

[itex] | | 3x + 1 | - x | = 2 [/itex]

## Homework Equations

[itex] |x| = x [/itex] if [itex]x \geq 0 [/itex]

[itex] |x| = -x [/itex] if [itex]x < 0 [/itex]

## The Attempt at a Solution

To start, I have never dealt with such an equation before, and am partly 'iffy' with absolute values in general, but, up until now in my calculus textbook (Calculus A First Course), I have been okay with them.

I would begin to think that there could be four possibilities: the inner absolute value could be greater than or equal to zero in combination with the outer absolute value being greater than or equal to zero, or it could be a combination of greater than or equal to zero and less than zero, the reverse of that, and both being less than zero.

And, there must be a method to see which of the four solved x-values are true.

Supposedly, if there are two absolute values in an equation, one must create the intervals of the domain and see which solved x-values exist in the interval you are solving in.

For example, I would create the following two inequalities and solve for x:

[itex]3x + 1 \geq 0[/itex]

[itex]3x \geq -1 [/itex]

[itex]x \geq -\frac{1}{3}[/itex]

[itex]3x + 1 - x \geq 0[/itex]

[itex]2x \geq -1[/itex]

[itex]x \geq -\frac{1}{2}[/itex]

Then, the corresponding 'less than' inequalities would be done.

[itex]3x + 1 < 0[/itex]

[itex]3x < -1[/itex]

[itex]x < -\frac{1}{3}[/itex]

Here, I am not certain as to using [itex]3x + 1[/itex] or [itex]-(3x + 1)[/itex]

[itex]3x + 1 - x < 0[/itex]

[itex]2x < -1[/itex]

[itex] x < -\frac{1}{2}[/itex]

So, I now have the domain,

[itex] x < -\frac{1}{3}, -\frac{1}{3} \leq x < -\frac{1}{2}, x \geq -\frac{1}{2}[/itex]

Now, I will do cases in each interval, where the nature of the absolute values will be decided by whether they are positive or negative in the interval. And, if the solved x-value exists in the interval, it is a solution.

__CASE 1__

[itex] x < -\frac{1}{3}[/itex]

[itex]3(-2) + 1 < 0[/itex]

(Here, I am not sure again to use the positive absolute value or negative.)

[itex]3(-2) + 1 - (-2) < 0[/itex]

Therefore,

[itex]-[-(3x + 1) - x] = 2[/itex]

[itex]-(-3x - 1 - x) = 2[/itex]

[itex]-(-4x - 1) = 2[/itex]

[itex]4x + 1 = 2[/itex]

[itex]x = \frac{1}{4}[/itex]

However, 1/4 does not exist in this interval.

Now, I would continue on, but it strongly appears that with this method I will not get the answers supplied in the textbook. So, any help on how to do this correctly would be much appreciated. (As a note, I was able to solve it and get the correct answers on first try [not through this method], but then I realized I had no way of proving why these two values of the four were correct and the other two were not.)

Much appreciation for any help!