1. ## |3x-1|<|-x+4|

Solve in set R
|3x-1|<|-x+4|

2. Originally Posted by lebanon
Solve in set R
|3x-1|<|-x+4|
first find intervals in which you will look

$\displaystyle 3x-1 = 0 \Rightarrow x = 1/3$
$\displaystyle -x +4 = \Rightarrow x = 4$

so you look in intervals $\displaystyle (-\infty, 1/3 ] \cup [1/3 , 4 ] \cup [4 , \infty)$ of the interval $\displaystyle (-\infty, \infty)$

can you continue ? show what have you done

3. Another way to do this: Look at the equation |3x- 1|= |-x+ 4|. Since the two absolute values are the same, either the numbers themselves are the same, 3x- 1= -x+ 4, or the differ only in sigh, 3x-1= -(-x+ 4)= x- 4.

The first equation, 3x- 1= -x+ 4 gives 4x= 5 or x= 5/4. The secon gives 2x= -3 or x= -3/2.

The point is that those two points are the only points at which "<" can change to ">" and vice versa.

The number -2 is less that -3/2 and putting x= -2 in the original equation, |3(-2)-1|= |-7|= 7> |-(-2)+4)|= |6|= 6 so |3x-1|> |-x+ 4| for all x< -3/2.

The number 0 is between -3/2 and 5/4. Putting x= 0 in the original equation, |3(0)-1|= |-1|= 1< |-(0)+ 4|= |4|= 4 so |3x-1|< |-x+ 4| for all x between -3/3 and 5/4.

The number 2 is greater than 5/4. Putting x= 2 in the original equation, |3(2)- 1|= |5|= 5> |-(2)+ 4|= |2|= 2 so |3x-1|> |-x+ 4| for all x greater than 5/4.

4. Originally Posted by lebanon
Solve in set R
|3x-1|<|-x+4|
you can square and get $\displaystyle 9x^2-6x+1<x^2-8x+16\implies8x^2+2x-15<0,$ i think you can continue.