$\begin{array}{l}
\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1 \\
\Rightarrow \frac{16}{x}-\frac{15}{x+1}=1 \\
\Rightarrow \frac{16 x+16-15 x}{x(x+1)}=1 \\
\Rightarrow \frac{x+16}{x^{2}+x}=1 \\
\Rightarrow x^{2}+x=x+16 \\
\Rightarrow x^{2}-16=0 \\
\Rightarrow(x+4)(x-4)=0 \\
\Rightarrow x+4=0 \text { or } x-4=0 \\
\Rightarrow x=-4 \text { or } x=4
\end{array}$

(Cross multiplication)

Hence, $-4$ and 4 are the roots of the given equation.