Question

How do you solve `absx> x - 1`?

Answer 1

When dealing with moduli it's often useful to split into cases where the enclosed value changes sign. For this example, that means examine the `x < 0` and `x >= 0` cases separately.

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Case (a) : `x < 0`

If `x < 0` then `|x| = -x` and the inequality takes the form

`-x > x - 1`

Add `x + 1` to both sides to get:

`2x < 1`

Divide both sides by `2` to get:

`x < 1/2`

Since we're dealing with the case `x < 0` this condition is already satisfied.

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Case (b) : `x >= 0`

If `x >= 0` then `|x| = x` and the inequality takes the form

`x > x - 1`

Subtract `x` from both sides to get:

`0 > -1`

...which is always true.

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Putting these two cases together, we find that the original inequality is satisfied for all `x in RR`.