Solving equations with mod So, I'm trying to solve the following equation using regular algebra, and I don't think I'm doing it right: $3x+5 = 1\pmod {11}$
I know the result is $x = 6$, but when I do regular algebra like the following, I do not get 6:
$3x=1  - 5\pmod{11}$
$x = \dfrac{(1 \pmod{11} - 5)} 3$
So, I figured that since $1 \pmod{11} = 1$ the equation becomes
$x = \dfrac{-4} 3$
Which is not 6! I am totally lost here and would appreciate any help......

The congruence relation 3x+5≡1 (mod 11)  is equivalent to 3x-11y= -4 where x,y ∈Z.
My solution to the problem may be found in the link to the PDF document: http://www.aespen.ca/AEnswers/1415545790.pdf.
The following image was created from the PDF document above.

