Find the roots of the given equation:

contexto.140

3 years ago

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

(Cross multiplication)

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