## College Algebra (6th Edition)

a. 0 b. $\{ 4 \}$
$\frac{x-2}{2x} + 1 = \frac{x+1}{x}$ 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 $:\{ 4 \}$