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 \(-2\sqrt6\).