rip brilliant community problems part 2

$\left(\dfrac{k+2}{3}\red{\times}\blue{1}\red{5}=\dfrac{k-4}{5}\blue{\times}\red{1}\blue{5}\right)$ $\Rightarrow \red{5k}\blue{+}\red{10}\blue{=}\red{3k}\blue{-}\red{12}$ $\Rightarrow \red{2}\blue{k}\red{=}\blue{2}\red{2}$ $\Rightarrow \boxed{\red{k}\blue{=}\red{-}\blue{1}\red{1}}$

I will miss community problems 😔 Rearranging, we get 5k+10=3k-12 Thus 2k=-22, k=-11

There's no place like Brilliant.

$\dfrac{k+2}{3}=\dfrac{k-4}{5}$

$5(k+2)=3(k-4)$

$5k+10=3k-12$

$2k=-22$

$\dfrac{2k}{2}=\dfrac{-22}{2}$

$\boxed{k=-11}$