Question

Solve 4y=32y/3 - y^2

Answer 1

In order to get rid of the fraction, I basically just multiplied everything by 3:
\[3(4y=32y/3-y ^{2})\]
which then gave me...
\[12y=32y-3y ^{2}\]
so, moving everything around, I got:
\[3y ^{2}+12y-32y\]
so, simplify... and you get:
\[3y ^{2}-20y=0\]
aaaand, solve for y!
\[y(3y-20)=0\]
\[y=0, y=20/3\]

let me know if you have any other questions! hope this helps! :)

Answer 2

thanks

Answer 3

no prob :)