\(\displaystyle gcd(36,49)=1\)

\(\displaystyle 49=36*1+13\)

\(\displaystyle 36=13*2+10\)

\(\displaystyle 13=10*1+3\)

\(\displaystyle 10=3*3+1\)

\(\displaystyle 3=1*3+0\)

Working backwards I obtain: \(\displaystyle 1=36*15-49*11\)

\(\displaystyle -11\equiv y \ \mobx{(mod 36)}\rightarrow -11\equiv 25 \ \mbox{(mod 36)}\)

\(\displaystyle x\equiv 22*25 \ \mbox{(mod 36)}\rightarrow x\equiv 10 \ \mbox{(mod 36)}\)

\(\displaystyle x_0=10\)

\(\displaystyle x=10+36t \ 0\leq t<1\)

\(\displaystyle x=10 \ \mbox{is the in-congruent solution}\)

Is this correct and if so, is there an easier way to do this?