SOLUTION: {{{1/2+((2x-1)/3)<=1}}} please help me sole this inequality in interval notation thank you note the ( ) are actually absolute value brackets

Algebra ->  Absolute-value -> SOLUTION: {{{1/2+((2x-1)/3)<=1}}} please help me sole this inequality in interval notation thank you note the ( ) are actually absolute value brackets      Log On


   

Question 439096: 1%2F2%2B%28%282x-1%29%2F3%29%3C=1
please help me sole this inequality in interval notation
thank you
note the ( ) are actually absolute value brackets
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
1/2+|2x-1/3)|<=1
1/2+|2x-1|/3<=1 (pull out denominator 3)
LCD=6
3+2|2x-1|<=6
solve for two possibilities: (2x-1)>=0 or (2x-1)<0
..
3+2(2x-1)<=6
3+4x-2<=6
4x>=5
x>=5/4
..
3+2(-2x+1)<=6
3-4x+2<=6
-4x>=1 (divide by (-4) and reverse inequality sign
x<=-1/4
..
Solution:
(-infinity,-1/4] U [5/4,infinity)