(x2 -x, y2 -3y+2) = (0, 0)
x2 -x = 0 or x(x-1)=0 or x = 0, 1
and y2 -3y+2 = 0 or (y-1) (y-*2) = 0 or y = 1, 2
Hence, x = 0, x = 1, y = 1, y = 2
If A = {a, b, c} and B = {p, q}, then find :
(i) A x B (ii) B x A (iii) A x A (iv) B x B (v) n (A x B) (vi) n (B x A) (vii) n (A x A) (viil) n (B x B).
(i) A X B = {a, b, c} x (p, q} = {{a, p}, {a, q}, (b, p}, {b, q}, {c, p}, {c, q}}
(ii) B x A = {p, q} x {a, b, c} = {{p, a, }.{p, b}.{p, c}.{q, a}.{q, b}.{q, c}}
(iii} A x A = {a, b, c} x {a, b, c} = {(a, a), (a, b), (a, c), (b,l, a), (b,, b), (b, c), (c a), (c, b), (c, c)}
(iv) B x B = {p.q} x {p.q} = {(p.p), (p.q), (p.q), (p.q)
(v) n(A x B) = n(A) x n(B) = 3 x 2 = 6
(vi) n(B xA) = n(B) x n(A) = 2 x 3 = 6
(vii) n(A xA) = n(A) x n(A) = 3 x 3 = 9
(viii) n(B x B) = n(B) x n(B) = 2 x 2 = 4
(2x, x + y) = (6, 2)
2x = 6 or x = 3
and x + y = 2 or 3+y = 2 or y = -1
Hence, x = 3, y = -1
A = {1, 2, 3}
The only even priome number is 2 and it does not lie between 10 a nd 27.
∴
If A = {1, 2}, B = {3, 4} and C ={x : x is even x divides 15}, that show that
A = {1, 2}, B = {3, 4}
C = {x : xis even and x divides 15}
Since no even number divides 15
∴