Question

How to solve1 /|2x-3| >1 ?

Answer 1

{x`epsilon`R | `1<x<2`, `x!=3/2`}

Explanation:

Since an inequality changes directions only when multiplied or divided by a negative, and since |2x-3| is always positive, multiply both sides of the relation by |2x-3|. The direction of the inequality remains the same:
`1/|2x-3|>1`
`color(white)(aaaiiaa)` `1 > |2x-3|` `color(white)(aaaiiaa)`Then, rearranging,

`color(white)(i)` `|2x-3|<1`
By the definition of absolute value, (2x-3) must be less than a distance of 1 from 0 on the number line. Two equations result:
`color(white)(aaaiiaa)` `2x-3<1` and `2x-3> -1`
Solving each separately,
`color(white)(aaaiiaaaaa)` `2x<4` `color(white)(aaaaaiaa)` `2x>2`
`color(white)(aaaiiaaaaaa)` `x<2` `color(white)(aaaaaaiaa)` `x>1`

The solution then is
`1<x<2` with the restriction that `x!=3/2` since that value would make the denominator zero in the original question.