Answer
a. 0
b. $\{ 4 \}$
Work Step by Step
$\frac{x-2}{2x} + 1 = \frac{x+1}{x}$
a. $x =0 $ makes the denominator zero. So, $x \ne 0.$
b. $\frac{x-2}{2x} + 1 = \frac{x+1}{x} ; x \ne 0;$
$\frac{x-2+2x}{2x} = \frac{x+1}{x} ; x \ne 0;$
$\frac{3x-2}{2x} = \frac{x+1}{x} ; x \ne 0;$
Multiply both sides by $2x$.
$2x(\frac{3x-2}{2x} ) =2x( \frac{x+1}{x} ); x \ne 0;$
$3x-2 = 2(x+1)$
$3x-2 = 2x+2$
$3x-2x = 2+2$
$x = 4$
$4$ is not the restricted value. So,
Solution set $:\{ 4 \}$
