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

1 Answer
Feb 23, 2018

{xepsilonR | 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<4color(white)(aaaaaiaa)2x>2
color(white)(aaaiiaaaaaa)x<2color(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.