Algebra

Answer: $2\sqrt6$ and $-2\sqrt6$

Solution:

Multiplying both sides by 3y gives:

$y^2 + 3y = 3y + 24$

Subtracting both sides of the new equation by 3y gives:

$y^2 = 24$

Isolating y gives:

$y = \pm \sqrt24$

The square root of 24 can be simplified into $2\sqrt6$. So y can be $2\sqrt6$ or $-2\sqrt6$.