Oh, wait!
Is it 1/3x ?
Is what "1/3x"? And do you mean 1/(3x) or (1/3)x?
To divide by 3x+1, first write [tex]x^2= x^2+ 0x+ 0.
How many times does "3x" divide into ? (1/3)x times, of course. But you are not done. Now subtract 3x+1 times (1/3)x from : . How many times does 3x divide into -(1/3)x? -1/9 times. And now you subtract (3x+1) times -1/9 from -(1/3)x+ 0: -(1/3)x- (-1/9)(3x+1)= -(1/3)x+ (1/3)x+ 1/9= 1/9.
3x+1 divides into (1/3)x- 1/9 times with remainder 1/9.