What is #-5/9+ 1 1/3#?

Basically, you find the LCM of the denominators. That of 3 and 9 being 9, you multiply the numerator of the fraction with the smaller denominator with whatever multiple of the denominator the LCM is.
#-5/9+1 1/3 =4/3-5/9 =(12-5)/9 =7/9#

Perhaps a simpler visualisation would be thus:

#-5/9 + 4/3*3/3 = -5/9+12/9 = (12-5)/9 = 7/9#

#1#. Start by converting #1##1/3# into an improper fraction.

#color(darkorange)1##color(teal)1/color(violet)3#

#=(color(violet)3color(darkorange)(xx1)color(white)(i)color(teal)(+1))/color(violet)3#

#=4/3#

#2#. Find the L.C.M. (lowest common multiple) between the denominators of the two fractions.

#|ul(9color(white)(X)3)#

#color(darkorange)3|ul(9color(white)(X)3)#
#color(white)(Xx)color(teal)3color(white)(X)color(violet)1#

L.C.M.#=color(darkorange)3xxcolor(teal)3xxcolor(violet)1=9#

#3#. Write out the problem with improper fractions. If the denominator of any of the fractions is not the L.C.M., #9#, multiply the numerator and denominator of the fraction by a number such that the denominator becomes the L.C.M., #9#. In your case, multiply the numerator and denominator of #4/3# by #3#.

#-5/9+4/3#

#=-5/9+(4color(red)(xx3))/(3color(red)(xx3))#

#=-5/9+12/9#

#4#. Now that both fractions have the same denominator, you can add them the numerators of the two fractions together.

#=(-5+12)/9#

#=color(green)(|bar(ul(color(white)(a/a)7/9color(white)(a/a)|)))#