15(y-4)-2(y-9)+5(y+6)=0

(15(2-y)-5(y+6))/(1-3y)=10

(x-4)(x-3)=(x-6)(x-5) (y-9)(y-3)= (y-4)(y-3)

(6y-5)/(2y)=7/9

Solve for x and y : (x, y ne 0) (10)/(2x+y)+(3)/(2x-y)=3, (15)/(2x+y)+(9)/(2x-y)=6

solve: (2) / (2x + y) - (1) / (x-2y) + (5) / (9) = 0 and (9) / (2x + y) - (6) / (x-2y ) + 4 = 0

If the circumference of the circle x^2+y^2+8x+8y-b=0 is bisected by the circle x^2+y^2=4 and the line 2x+y=1 and having minimum possible radius is 5x^2+5y^2+18 x+6y-5=0 5x^2+5y^2+9x+8y-15=0 5x^2+5y^2+4x+9y-5=0 5x^2+5y^2-4x-2y-18=0

Find value of x and y: (5x-4y+8=0),(7x+6y-9=0)