`x^3-3x y^2=18 3x^2y-y^3=26 ` If `x , y in N`

2x+3y=18;x-2y=2

(x^3+3x y^2)dx+(y^3+3x^2y)dy=0

From the sum of x^2+3y^2-6x y ,2x^2-y^2+8x y , y^2+8 and x^2-3x y\ subtract -3x^2+4y^2-x y+x-y+3.

From the sum of x^2+3y^2-6x y ,2x^2-y^2+8x y , y^2+8 and x^2-3x y\ subtract -3x^2+4y^2-x y+x-y+3.

x + y = 3, 4 x - 3y = 26 .

Subtract the sum of - 3x^3y^2 + 2x^2 y^3 and - 3x^2y^3 - 5y^4 " from " x^4 + x^3y^2 + x^2y^3 + y^4

3x +4y le 18, 2x +3y ge 3 and x, y ge 0 is

Solve for (x - 1)^(2) and (y + 3)^(2) , 2x^(2) - 5y^(2) - x - 27y - 26 = 3(x + y + 5) and 4x^(2) - 3y^(2) - 2xy + 2x - 32y - 16 = (x - y + 4)^(2) .

If 3x + y i= x + 2y , then 2x - y =