How do I prove the equivalence of these two congruences?

Question

I have $7x\equiv 1\pmod8$.How do I prove it is equivalent to $x\equiv 7\pmod8$?

I have no idea to start on this question.Thanks for any reply..

Answer 1

$7x\equiv1({\rm mod}8)\iff-x\equiv1({\rm mod}8)\iff x\equiv-1\equiv7({\rm mod}8)$

Answer 2

You need to find the multiplicative inverse of the coefficient in front of $x$ modulo a given number. You can do this by repeating the Euclidean Algorithm.

If such a number doesn't exist the equation doesn't have a solution (given that in $ax = b \pmod c$ we have $gcd(a,b,c) = 1$)

Answer 3

The solution can be done in this way.$7x\equiv1\pmod8$ this can be written in the form

$8x-x\equiv8-7\pmod8$ and we get $-x\equiv-7\pmod8$.Then multiplying $-1$ on both sides we get

$x\equiv7\pmod8$