Value of x makes denominator equal to 0: $ S=\{ \}$ (impossible) $0x =a \not = 0: S=\{ \}$ (impossible) $0x = 0: S=\mathbb R $ (indeterminate)

Got it !

2x-$\frac{2x}{9}=\frac{1}{9}(1x-\frac{3}{2})$	$ 0x=-3\qquad;\qquad S =\{\} $
$\frac{5x}{18}-\frac{4x-3}{8}=\frac{9-2x}{9}$	$0x=45\ \qquad;\qquad S =\{\} $
$\frac{3x}{2}-\frac{2x}{3}=5(\frac{x}{6}+1)-5$	$0x=0 \qquad;\qquad S =\mathbb R $
$\frac{2}{x+1}+\frac{5}{x-1}=\frac{10}{x^2-1}$	$x=1 \qquad;\qquad S =\{\} $
$\frac{2x-3}{5}+\frac{x}{2}= \frac{3(3x-2)}{10}$	$0x=0 \qquad;\qquad S =\mathbb R $
$\frac{4}{x}+\frac{x}{x+1}=\frac{x^2}{x^2+x}\frac{}{}$	$x=-1 \qquad;\qquad S =\{\} $