" 5.हल Do it : "quad (x)/(2)+(2y)/(3)=-1" And "x-(y)/(2)=3

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

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

Solve for x and y 2/(3x+2y)+3/(3x-2y)=17/5 5/(3x+2y)+1/(3x-2y)=2

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 x^2=y^3 , then prove that (x/y)^(3/2)+(y/x)^-(2/3)=x^(1/2)+y^(1/3) .

Solve for x and y 2/(3x+2y)+3/(3x-2y)=17/5,5/(3x+2y)+1/(3x+2y)=6/5

(3) x+2y=-1;2x-3y=12

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