1. ## inequalities

i dont know but i just cant seem to understand how the inequalities have been obtained

given that 0<lx-2l<1,
i get that it becomes -1<x-2<1

but the book went on to say that
-1<x-2<3
2<3x-1<8
thus l 3x-1 l>2

how is the LHS of the inequalities obtained? namely that it is -1 and 2..

2. Originally Posted by alexandrabel90
i dont know but i just cant seem to understand how the inequalities have been obtained

given that 0<lx-2l<1,
i get that it becomes -1<x-2<1

but the book went on to say that
-1<x-2<3
2<3x-1<8
thus l 3x-1 l>2

how is the LHS of the inequalities obtained? namely that it is -1 and 2..
Dear alexandrabel90,

$\displaystyle -1<x-2<1\mbox{~and~}1<3\Rightarrow -1<x-2<3$ --------(1)

$\displaystyle -1<x-2<1$

$\displaystyle \Rightarrow -3<3x-6<3$

$\displaystyle \Rightarrow -3+5<3x-6+5<3+5$

$\displaystyle \Rightarrow 2<3x-1<8$ --------(2)

Since, $\displaystyle 3x-2>2$; 3x-2 is a positive number. Therefore, $\displaystyle \mid 3x-2\mid>2$ ------------(3)

Hope you understood.

3. A slightly different way to do it: from 0< |x- 2|< 1 is the same as -1< x- 2< 1 (with the stipulation that $\displaystyle x- 2\ne 0$! That is -1< x-2< 0 or 0< x-2< 1) Adding 2 to each part, 1< x< 3 (with the added stipulation that x is not 2: 1< x< 2 or 2< x< 3). Multiply by 3: 3< 3x< 9 (with the added stipulation that 3x is not 6). Subtracting 1, 2< 3x-1< 8 (with the added stipulation that 3x-1 is not 5: 2< 3x- 1< 5 or 5< 3x- 1< 8. But since all possible values of 3x-1 are positive and larger than 2, it is true that, for all x satisfying 0< |x- 2|< 1, 2< |3x-1|