Re: simultanous equaltions

you can solve it substantially as you would an ordinary equation with a few caveats

a) negative coefficients have to be converted to their GF(7) equivalent

b) addition and subtraction are done mod 7 obviously

c) you have to be careful when "dividing". What you need to do is multiply by the multiplicative inverse of the number you want to divide by.

suppose you had $3x = 5$

with reals or rationals you'd just divide by 3 and say $x=\dfrac 5 3$

with GF(7) you have to note that the multiplicative inverse of 3 is 5, as $3 \cdot 5 = 15 = 1 \pmod{7}$

and then

$5 \cdot 3 x = 5 \cdot 5$

$x = 25 \pmod{7} = 4$

if you pay attention to those 3 rules you can solve it the way you usually do.