Verified answer

[tex]\mathbb{SOLUTION:}[/tex]

Reduce the fraction to the lowest term.

[tex]\quad\quad\quad\quad\quad\begin{aligned} \tt{\frac{2k}{3}} & = \tt{\frac{4k^{2}}{30}} \\ \tt{\frac{2k}{3}} & = \tt{\frac{2k^{2}}{15}} \end{aligned}[/tex]

Multiply the 15 on both sides of the equation by common denominator.

[tex]\quad\quad\quad\quad\begin{aligned} \tt{\frac{2k \times 15}{3}} & = \tt{\frac{2k^{2} \times 15}{15}} \\ \tt{10k} & = \tt{\frac{2k^{2} \times \bcancel{15}}{\bcancel{15}}} \\ \tt{10k} & = \tt{2k^{2}} \end{aligned}[/tex]

Reduce the common factor on both sides of the eqution, arrange the terms and factored out.

[tex]\quad\quad\quad\quad\quad\begin{aligned} \tt{10k} & = \tt{2k^{2}} \\ \tt{5k} & = \tt{k^{2}} \\ \tt{5k - k^{2}} & = \tt{0} \\ \tt{k(5 - k)} & = \tt{0} \end{aligned}[/tex]

Then, lets apply to the ZPP or the Zero Product Property and solve for the k.

[tex]\quad\quad\quad\quad\quad\begin{gathered} \tt{k(5 - k)} = \tt{0} \\ \tt{k = 0} \: \tt{or} \: \tt{5 - k = 0} \\ \tt{k = 0} \: \tt{or} \: \tt{-k = -5} \\ \boxed{\tt{k = 0} \: \tt{or} \: \tt{k = 5}} \end{gathered}[/tex]

Hence, the solutions are k = 0, k = 5