Answer
a. Restriction values $: 2$
b. Solution set $:$ Empty set $:\{ \}$
Work Step by Step
$\frac{2}{x-2}=\frac{x}{x-2} - 2$
a. $x =2 $ makes the denominator zero. So, 2 is the restricted value for $x$.
b. $\frac{2}{x-2}=\frac{x}{x-2} - 2 ; x \ne 2;$
$\frac{2}{x-2}=\frac{x-2(x-2)}{x-2} ; x \ne 2;$
$\frac{2}{x-2}=\frac{x-2x+4}{x-2} ; x \ne 2;$
$\frac{2}{x-2}=\frac{-x+4}{x-2} ; x \ne 2;$
Multiply both sides by $(x-2)$ to clear fraction part.
$(x-2)(\frac{2}{x-2})=(x-2)(\frac{-x+4}{x-2}) ; x \ne 2;$
$2=-x+4$
$x=4-2$
$x= 2$
$2$ is the restricted value for $x$. So, this equation contains No solution.
Solution set $:$ Empty set $:\{ \}$
