Find the in-congruent solutions to $\displaystyle 49x\equiv 22 \ \mbox{(mod 36)}$

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

$\displaystyle 49=36*1+13$

$\displaystyle 36=13*2+10$

$\displaystyle 13=10*1+3$

$\displaystyle 10=3*3+1$

$\displaystyle 3=1*3+0$

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?