Question 13315
{{{(2/3)(5y-4)-(3/5)(2y+8)=0}}}


ok, first thing...get rid of the fractions. Do this by multiplying all terms by 15..the lowest common multiple of 3 and 5 (the 2 denominators)...


{{{15(2/3)(5y-4)-15(3/5)(2y+8)=15*0}}}
{{{5(2)(5y-4)-3(3)(2y+8)=0}}}
{{{10(5y-4)-9(2y+8)=0}}}
50y - 40 -18y - 72 = 0


now just collect like terms together...
32y - 112 = 0
32y = 112
y = 112/32
--> y = 3.5


jon