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

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

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

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

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

Solve by matrix method (2/x) + (3/y) + (10/z) =4, (4/x) - (6/y) + (5/z) =1 , (6/x) + (9/y) - (20/z) =2 and x,y, z != 0

The equation of the circle passing through the point of intersection of the circle x^(2)+y^(2)=4 and the line 2x+y=1 and having minimum possible radius is (a) 5x^(2)+5y^(2)+18x+6y-5=0(b)5x^(2)+5y^(2)+9x+8y-15=0(c)5x^(2)+5y^(2)+4x+9y-5=0(c)5x^(2)+5y^(2)-4x-9y-5=0(c)5x^(2)+5y^(2)-4x-2y-18=0