Answer
a. Restriction values $: -1$
b. Solution set $:$ Empty set $:\{ \}$
Work Step by Step
$\frac{8x}{x+1} = 4 - \frac{8}{x+1}$
a. $x =-1 $ makes the denominator zero. So, $x \ne -1$
b. $\frac{8x}{x+1} = 4 - \frac{8}{x+1}; x \ne -1;$
$\frac{8x}{x+1} = \frac{4(x+1)-8}{x+1}; x \ne -1;$
$\frac{8x}{x+1} = \frac{4x+4-8}{x+1}; x \ne -1;$
$\frac{8x}{x+1} = \frac{4x-4}{x+1}; x \ne -1;$
Multiply both sides by $(x+1)$ to clear fraction.
$(x+1)(\frac{8x}{x+1} )= (x+1)(\frac{4x-4}{x+1}); x \ne -1;$
$8x = 4x-4$
$8x-4x=-4$
$4x=-4$
$x=\frac{-4}{4}$
$x= -1$
But we already told that if $x= -1$ , it makes the denominator zero. Therefore, this equation contains no solution.
Solution set $:$ Empty set $:\{ \}$
