#### Answer

a. $1$
b. $\{ 2 \}$

#### Work Step by Step

$\frac{3}{2x-2} + \frac{1}{2} = \frac{2}{x-1}$
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 : $\{ 2 \}$