SOLUTION: The problem is |3x+2| > 4. This is what I tried: 3x+2 > 4 or -3x-2>-4 3x > 2 or -3x>-2 x > 2/3 or x<2/3 Is this answer right? Also how do I graph it?

Algebra ->  Inequalities -> SOLUTION: The problem is |3x+2| > 4. This is what I tried: 3x+2 > 4 or -3x-2>-4 3x > 2 or -3x>-2 x > 2/3 or x<2/3 Is this answer right? Also how do I graph it?      Log On


   



Question 984547: The problem is |3x+2| > 4. This is what I tried:
3x+2 > 4 or -3x-2>-4
3x > 2 or -3x>-2
x > 2/3 or x<2/3
Is this answer right? Also how do I graph it?

Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The problem is |3x+2| > 4. This is what I tried:
3x+2 > 4
-----------------
or -3x-2>-4 **************** Not -4. -3x-2 > 4
---------------
3x > 2 or -3x>-2
x > 2/3 or x<2/3
Is this answer right? Also how do I graph it?

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
No your mistake is on the second line.  The -4 should be a +4.
You changed the sign of both the expression and the 4.  That's wrong.
You only change the sign on ONE side, not both. The 3x+2 > 4 is correct,
and x > 2/3 is correct.

Here is the correction:

The problem is |3x+2| > 4. This is what I tried:
3x+2 > 4 or -3x-2>4         
3x > 2 or -3x>6
x > 2/3 or x < -2

What's between a pair of absolute value bars can either be what's there or the
negative of what's there because |5| and |-5| are the same thing> So
the 5 can either have an understood positive sign before it or a written
negative sign before it.  But the right side (the number) doesn't change signs.

To graph that draw a numberline and mark -2 and 2/3 with open circles.  2/3 is
located two-thirds of the way between 0 and 1:  
                            
<==============o----------------o=============>
-4     -3     -2     -1     0     1     2    3   

The interval notation for that is 



Edwin