SOLUTION: Solve {{{abs(5x-3)>=12}}}

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Question 199400: Solve abs%285x-3%29%3E=12
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve absolute value 5x-3> or equal to 12
---------
Do it the same way as you do with an equal sign.
5x-3>=12
Add 3
5x>=15
Divide by 5
x>=3
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Ooops, I didn't notice the absolute value. Use Edwin's solution.
My solution is not incorrect, it's incomplete.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Alan's solution is incorrect!
Edwin's solution:

Solve: abs%285x-3%29%3E=12

Rules for removing absolute value bars:

1. For cases of abs%28EXPRESSION%29%3CA where A is positive:

Rewrite as

-A%3CEXPRESSION%3CA 

2. For cases of abs%28EXPRESSION%29%3C=A

Rewrite as

-A%3CEXPRESSION%3C=A 

3. For cases of abs%28EXPRESSION%29%3EA where A is positive:

Rewrite as

EXPRESSION%3C-A OR EXPRESSION%3EA

4. For cases of abs%28EXPRESSION%29%3E=A

Rewrite as

EXPRESSION%3C=-A OR EXPRESSION%3E=A


abs%285x-3%29%3E=12

This is case 4, so to remove the absolute value
bars:

abs%285x-3%29%3E=12

Rewrte that as

5x-3%3C=-12 OR 5x-3%3E=12

Solve each

5x%3C=-9 OR 5x%3E=15

x%3C=-9%2F5 OR x%3E=3

x%3C=-1%264%2F5 OR x%3E=3

Mark the points -1%264%2F5 and 3 on the number line.
Notice that to plot -1%264%2F5 you mark a point 4%2F5
of the way between -1 and -2 



We mark them with solid circles because the problem has %22%22%3C=%22%22
and/or %22%22%3E=%22%22.  If it had been just %22%22%3C%22%22 and/or %22%22%3E%22%22
you would have left the circles open


 ------------@-----------------------@----------
 -4   -3   -2   -1    0    1    2    3    4    5   
     
For the x%3C=-1%264%2F5 part we shade left of -1%264%2F5


<============@-----------------------@----------
 -4   -3   -2   -1    0    1    2    3    4    5

And since the solution is that OR x%3E=3 we shade
right of 3:


<============@-----------------------@===========>
 -4   -3   -2   -1    0    1    2    3    4    5

Now for the interval notation, the left shaded part is

(-infinity,--1%264%2F5]

and we union that with the right shaded part [3,oo)

and the final answer in interval notation is

        (-infinity,-1%264%2F5] U [3,infinity)

Edwin