SOLUTION: 6x-3 divided by 3 in absolute value greater or equal to 7 dont know how to tell which method to use the or method or other

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Question 715875: 6x-3 divided by 3 in absolute value greater or equal to 7 dont know how to tell which method to use the or method or other
Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
abs%28%286x+-+3%29%2F3%29 ≧ 7

Simplify by factoring out 3 in the numerator:

abs%283%282x+-+1%29%2F3%29 ≧ 7

Then cancel the 3's

abs%28cross%283%29%282x+-+1%29%2Fcross%283%29%29 ≧ 7

|2x - 1| ≧ 7

There are four rules for removing absolute value bars in
absolute value inequalities:

If A is positive:

1.  |EXPRESSION| < A becomes 

-A < EXPRESSION < A


2.  |EXPRESSION| ≦ A becomes 

-A ≦ EXPRESSION ≦ A


3.  |EXPRESSION| > A becomes

EXPRESSION < -A  OR EXPRESSION > A 


4.  |EXPRESSION| ≧ A becomes

EXPRESSION ≦ -A  OR EXPRESSION ≧ A

-----------------------------------------

        |2x - 1| ≧ 7

is rule 4 above where EXPRESSION = 2x - 1 and A = 7

    |2x - 1| ≧ 7 becomes

2x - 1 ≦ -7  OR  2x - 1 ≧ 7

Solve each for x:

    2x ≦ -6 OR  2x ≧ 8  

Divide through by 2

     x ≦ -3 OR  x ≧ 4

Draw the graph on a number line:

<=========●--------------------●===========>
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8 

Then the interval notation for this is an abbreviation 
for the above graph:

                   (-∞,-3] U [4,∞)

Edwin