Geometry Help!!!!!!!!!!
posted by Nina .
If the sides of a triangle have measurements 3x + 4 , 6x  1, and 8x + 2, find all possible values of x.
I also need help with this question please

All sides must be greater of zero.
3x+4 always>0
8x+2 always>0
Only side: 6x1 can be negative.
That's why:
6x1>0
6x>1 Divide both sides with 6
x>1/6 
Thanks anyway but that isn't one of the choices the answer is x> 3/11 i figured it out thanks for trying :)

Anonymus solution not completely.
All sides must be greater of zero:
3x+4>0
3x> 4 Divide with 3
x> 4/3
6x1>0
6x>1 Divide with 6
x>1/6
8x+2>0
8x> 2 Divide with 8
x> 2/8
x> 1/4
Least of that numbers is 4/3= 1.3333
x> 4/3 is solution 
x>3/11 is also > 4/3
All x> 4/3 is choices 
The condition that all sides have to be positive, as Anonymous used, is not sufficient.
In any triangle the sum of 2 sides must be greater than the third side, so
3x+4 + 6x1 > 8x + 2 > x > 1
AND
3x+4 + 8x + 2 > 6x1 > x > 7/5
AND
8x + 2 + 6x  1 > 3x+4 > x > 3/11
the intersection of all three conditions is
x > 3/11
To show that Anonymous is incorrect, pick a value of x between his/her answer of 1/6 and mine of 3/11
e.g. x = 11/50
3x+4 > 4.66
6x1 > .32
8x+2 > .24 , contradiction
I have a value of x > 1/6 which did not work.
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