Question

$\frac{16}{x}-1=\frac{15}{x+1};x\ne 0,-1$,one of the value of x is

Answer 1

$\frac{16}{x}-1=\frac{15}{x+1}$
$\frac{16-x}{x}=\frac{15}{x+1}$
$16x+16-x^2-x=15x$
$x^2=16$
$x=\pm 4$