[" 0."(2)/(x)+(5)/(y)=1" V "],[(60)/(x)+(40)/(y)=19,x!=0,y!=0]

Solve the following system of equations: (2)/(x)+(5)/(y)=1,quad (60)/(x)+(40)/(y)=19,x!=0,y!=0

Solve the following system of equations: 2/x+5/y=1,\ \ \ \ (60)/x+(40)/y=19 ,\ x!=0,\ y!=0

(3) / (x) - (1) / (y) + 9 = 0 (2) / (x) + (3) / (y) = 5, (! = 0, y! = 0)

(xy)/(x-y)=(1)/(2);(xy)/(x+y)=(1)/(5) wherex +y!=0;x-y!=0

Solve : (3)/(x)-(2)/(y)=0 and (2)/(x)+(5)/(y)=19 . Hence , find 'a' if y=ax+3

Solve for x and y:(xy)/(x+y)=(1)/(5),x+y!=0;(xy)/(x-y)=(1)/(7),x-y!=0

If y!=0 then the number of values of the pair "(x,y)" such that x+y+(x)/(v)=(1)/(2) and " (x+y)(x)/(y)=-(1)/(2) is

The locus of a point "P" ,if the join of the points (2,3) and (-1,5) subtends right angle at "P" is x^(2)+y^(2)-x-8y+13=0 x^(2)-y^(2)-x+8y+3=0 x^(2)+y^(2)-4x-4y=0,(x,y)!=(0,4)&(4,0) x^(2)+y^(2)-x-8y+13=0,(x,y)!=(2,3)&(-1,5)

(i) 15x^(2) - 8x + 1 = 0 (ii) 6y^(2) - 19y + 10 = 0