College Algebra (6th Edition)

a. $1$ b. $\{ 2 \}$
$\frac{3}{2x-2} + \frac{1}{2} = \frac{2}{x-1}$ Takeout common factor in the denominator. $\frac{3}{2(x-1)} + \frac{1}{2} = \frac{2}{x-1}$ a. $x=1$, makes the denominator zero,$x=1$ is the restricted value. b. $\frac{3}{2(x-1)} + \frac{1}{2} = \frac{2}{x-1} ; x \ne 1;$ Take LCD at the left hand side. $\frac{3+x-1}{2(x-1)} = \frac{2}{x-1} ; x \ne 1;$ $\frac{2+x}{2(x-1)} = \frac{2}{x-1} ; x \ne 1;$ Multiply both sides by $2(x-1)$. $2(x-1)(\frac{2+x}{2(x-1)}) = 2(x-1)(\frac{2}{x-1}) ; x \ne 1;$ $2+x = 4$ $x = 4 - 2$ $x = 2$ $2$ is not the restricted value. Solution Set : $\{ 2 \}$