" 3.यदि "(4x^(2)-3y^(2)):(2x^(2)+5y^(2))=12:19" हो,तो "x:yquad " 7."15:19

If (4x^2-3y^2):(2x^2+5y^2)=12:19 then x:y=

(1)/(3)x+5y=13,2x+(1)/(2)y=19

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) .

3x^(2) + 8x +4=0, 4y^(2)-19y + 12=0

Solve for x and y:3x^(2)-5y^(2)=7,3xy-4y^(2)=2

The radical axis of the circles x^(2) + y^(2) + 4x + 8y + 19 = 0 , x^(2) + y^(2) + 8x + 4y + 19 = 0 is

If (5x+2y) : (10x+3y) = 5:9 , then (2x^2+3y^2 ):(4x^2+9y^2) =? यदि (5x+2y) : (10x+3y) = 5:9 , है, तो (2x^2+3y^2 ): (4x^2+9y^2) =?

The radical centre of the circles x^(2)+y^(2)=9 , x^(2)+y^(2)-2x-2y-5=0 , x^(2)+y^(2)+4x+6y-19=0 is A) (0,0) (B) (1,1) (C) (2,2) (D) (3,3)