In solving this equation, it would be useful for us to group like terms on opposite sides of the equation. In this case this would entail moving all terms which contain our variable #y# to one side of the equation, and moving all of our terms not containing #y# to the other. This is performed below:

#4/5 y + 6 = 2/5 y -7 => 4/5 y - 2/5 y + 6 - 6 = 2/5 y - 2/5 y -7 - 6# (we subtract #2/5 y# and 6 from each side)

#=> 2/5 y = -13# (we simplify each side by combining like terms, and find that #2/5 y - 2/5 y# cancels out, as does #6-6#, while #4/5 y - 2/5 y = 2/5 y # and #-7 - 6 = -13#

Thus, we know #2/5 y = -13#. From here, to get #y# by itself, we multiply both sides by #5/2#, cancelling out the #2/5# on the left:

#2/5 y = -13 => (5/2)(2/5) y = (5/2)(-13) => (10/10) y = -65/2 => y = -32 1/2 = -32.5#

And we have obtained our solution.