" 3."(x)/(3)+(y)/(2)=3,x-2y=2

Solve the following system of equations: (2)/(3x+2y)+(3)/(3x-2y)=(17)/(5),quad (5)/(3x+2y)+(1)/(3x-2y)=2

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

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

((x +y)^(3) + (x-y)^(3))/(2) -y (3x^(2) + y^(2)) = ______

Solve : (x+y)^(2//3)+2(x-y)^(2//3)=3(x^2-y^2)^(1//3) and 3x-2y=13 .

The triangle formed by 2x+3y+7=0 and the pair (2x+3y)^(2)-3(3x-2y)^(2)=0 is

The equation of the transvers and conjugate axes of a hyperbola are,respectively,x+2y-3=0 and 2x-y+4=0, and their respective lengths are sqrt(2) and 2sqrt(3) .The equation of the hyperbola is (2)/(5)(x+2y-3)^(2)-(3)/(5)(2x-y+4)^(2)=1(2)/(5)(x-y-3)^(2)-3(x+2y-3)^(2)=12(x+2y-3)^(2)-3(2x-y+4)^(2)=1

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

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