What are the natural numbers x if y is integer? y=(6x-8)/(2x+1)

We'll write the numerator: 6x-8=2x+4x
-9+1

y=(6x-8)/(2x+1)=[(2x+1)/(2x+1)]+[(4x-9)/(2x+1)]

y=1+[(4x-9)/(2x+1)] 

We'll
try to do the same with the ratio
[(4x-9)/(2x+1)]=[(2x+1+2x-1-9)/(2x+1)]=

[(2x+1)/(2x+1)]+[(2x-10)/(2x+1)]=1+[(2x-10)/(2x+1)] 

So y
= 1+1+[(2x-10)/(2x+1)]

We'll follow the same
steps:

[(2x-10)/(2x+1)].

[(2x-10)/(2x+1)]=
[(2x+1-1-10)/(2x+1)]=1- [11/(2x+1)] 

y=1+1+1-
[11/(2x+1)]

y=3-
[11/(2x+1)]

If y is integer, the fraction [11/(2x+1)] has
to be also an integer number. For this reason, (2x+1) has to be the divisor of the
number 11. So, (2x+1) could be:+1,-1,+11,-11.

Now, we'll
put (2x+1) = 1

2x=0, x=0 and is a natural number, so it
follows the
constraint

y=(6*0-8)/(2*0+1)

y=-8/1

y=-8

(2x+1)
= -1

2x=-2

x=-1, but "-1" is
not a natural number, so (2x+1) is different
from -1 

(2x+1) =
11

2x=10

x=5 and is a natural
number.

y=(6*5-8)/(2*5+1)

y=22/11

y=2 

(2x+1)
= -11

2x=-12

x=-6 and is not a
natural number.

The natural values of x, that
makes the ratio y integer, are: {0 ; 5}.