First, multiply the #y/3# term by the appropriate form of #1# to create a common denominator with the other #y# fraction and the add the fractions:

#y/6 + (2/2 xx y/3) = 5/8#

#y/6 + (2 xx y)/(2 xx 3) = 5/8#

#y/6 + (2y)/6 = 5/8#

#(y + 2y)/6 = 5/8#

#(1y + 2y)/6 = 5/8#

#((1 + 2)y)/6 = 5/8#

#(3y)/6 = 5/8#

#(3y)/(3 xx 2) = 5/8#

#(color(red)(cancel(color(black)(3)))y)/(color(red)(cancel(color(black)(3))) xx 2) = 5/8#

#y/2 = 5/8#

Now, multiply each side of the equation by #color(red)(2)# to solve for #y# while keeping the equation balanced:

#color(red)(2) xx y/2 = color(red)(2) xx 5/8#

#cancel(color(red)(2)) xx y/color(red)(cancel(color(black)(2))) = color(red)(2) xx 5/((2 xx 4))#

#y = cancel(color(red)(2)) xx 5/((color(red)(cancel(color(black)(2))) xx 4))#

#y = 5/4#

Or

#y = 1.25#