A = {x : $x^2$ = 9 and 2x = 3} = {x : (x = 3 or x = –3} and x = $ \large\frac{3}{2}$ }

No x satisfies both the conditions. So there is no x in A i.e., A = $ \phi $

B = {x : $x^2$ – 5x + 6 = 0, 2x = 6}

= {x : (x – 3) (x – 2) = 0, x = 3}

= {x : (x = 3 or x = 2) = 0 and x = 3}

Only x = 3 satisfies both the predicates. So x has only one value 3.

$ \therefore $ B = {3}

C = {x : $ x^2$ – 4x + 3 = 0}

= {x : (x – 3) (x – 1) = 0}

= {x : x = 3 or x = 1}

2 and 1 both satisfy the statement x = 3 or x = 1

$\therefore $ C = {1 , 3}

Similarly, D = {-5, 5} and E = {1, 3}

Thus A is an empty set; B is a singleton set; C, D, E is pair sets.

Also, C = E