I always do these kind of stuff the way I learned it back where I was younger.
So, #55 5/9=((9xx55)+5)/9=(495+5)/9=500/9# and
#7 1/6=((6xx7)+1)/6=(42+1)/6=43/6#
Then the fun part of the division of two or more fractions, which is simply the numerator multiplied (times or #xx#) by the reciprocal of the denominator. Let's say #color(red)D# is the denominator, its #color(blue) (reciprocal)# will be #color(blue)(1/D)#. You can replace #color(red)D# by whatever number you want if letters bother you. Let's say #color(red)D=2#, its #color(blue) (reciprocal)# will be #color(blue)(1/D)=color(blue)(1/2)#.
So, our problem becomes simply
#55 5/9-:7 1/6=500/9-:43/6=(500/9)/(43/6)=500/(3cancel9) xx (2cancel6)/43=500/ 3xx2/43=1000/129#
Another why to find out what #55 5/9# is equal to is to tell yourself that there is an addition between #55# and #5/9#, which means,
#55 5/9=55+5/9=(495+5)/9=500/9# I used the common denominator (LCD)
Same thing for #7 1/6 =>7 1/6=7+1/6=(42+1)/6=43/6#
P.S. #color(blue) (RECIPROCAL)# is what some people often call #color(green) (INVERSE)# but they are indeed really different. Let's say we have the number #2#, its #color(blue) (RECIPROCAL)# is #color(blue)(1/2)# #color(red) (but)# its #color(green) (INVERSE)# is #color(green)(-2)#. So, the #color(green) (INVERSE)# of a "number" is just its #color(green) (OPPOSITE)#.
I'm talking about numbers here and not functions!
Hope this was helpful :)
