Solve each of the following quadratic equations:

$\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1$

$\Rightarrow \frac{16}{x}-\frac{15}{x+1}=1$

$\Rightarrow \frac{16 x+16-15 x}{x(x+1)}=1$

$\Rightarrow \frac{x+16}{x^{2}+x}=1$

$\Rightarrow x^{2}+x=x+16 \quad$ (Cross multiplication)

$\Rightarrow x^{2}-16=0$

$\Rightarrow(x+4)(x-4)=0$

$\Rightarrow x+4=0$ or $x-4=0$

$\Rightarrow x=-4$ or $x=4$

Hence, −4 and 4 are the roots of the given equation.