`(x + 2) (x + 1) = (x -2) (x -3)`
` (y+3)(y+2)=(y-1)(y-2)`

(x) / (2) + (y) / (3) = 1 (x) / (3) + (y) / (2) = 1

if (x_ (1), y_ (1)), (x_ (2), y_ (2)), (x_ (3), y_ (3)) are vertices equilateral triangle such that (x_ (1) -2) ^ (2) + (y_ (1) -3) ^ (2) = (x_ (2) -2) ^ (2) + (y_ (2) -3) ^ (2) = (x_ (3) - 2) ^ (2) + (y_ (3) -3) ^ (2) then x_ (1) + x_ (2) + x_ (3) +2 (y_ (1) + y_ (2) + y_ (3) ))

The solution of 2 (y + 3) - x y(dy)/(dx) = 0 with y =-2, when x =1 is :a) (y + 3) = x ^(2) b) x ^(2) (y + 3) = 1 c) x ^(4) (y + 3) = 1 d) x ^(2) ( y +3) ^(2) = e ^(y +2)

Find each of the following products: (i) (x - 4)(x - 4) (ii) (2x - 3y)(2x - 3y) (iii) ((3)/(4) x - (5)/(6) y) ((3)/(4)x - (5)/(6) y) (iv) (x - (3)/(x)) (x - (3)/(x)) (v) ((1)/(3) x^(2) - 9) ((1)/(3) x^(2) - 9) (vi) ((1)/(2) y^(2) - (1)/(3) y) ((1)/(2) y^(2) - (1)/(3) y)

If (2x + y + 2)/(5) = ( 3x - y + 1)/( 3 ) = ( 2 x + 2y + 1 )/(6) then

The factors of x^(3)-1+y^(3)+3xy are (a) (x-1+y)(x^(2)+1+y^(2)+x+y-xy)( b) (x+y+1)(x^(2)+y^(2)+1-xy-x-y)( c) (x-1+y)(x^(2)-1-y^(2)+x+y+xy)(d)3(x+y-1)(x^(2)+y^(2)-1)