Answer
The equation has two solutions, 1 and 3.
Work Step by Step
$\frac{8}{x+3}=\frac{1}{x}+1$
$\frac{8x}{x(x+3)}=\frac{x+3}{x(x+3)}+1$
$\frac{8x}{x^2+3x}-\frac{x+3}{x^2+3x}=1$
$\frac{8x-x-3}{x^2+3x}=1$
$7x-3=x^2+3x$
$x^2-4x+3=0$
$x^2-3x-x+3=0$
$x(x-3)-(x-3)=0$
$(x-1)(x-3)=0$
$x-1=0$ or $x-3=0$
$x=1$ or $x=3$
Check:
$8/(1+3)=8/4=2; 1/1+1=2$
$8/(3+3)=8/6=4/3; 1/3+1=4/3$
The equation has two solutions, 1 and 3.
