## College Algebra (6th Edition)

a. Restriction values $: 2$ b. Solution set $:$ Empty set $:\{ \}$
$\frac{2}{x-2}=\frac{x}{x-2} - 2$ 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 $:\{ \}$