((5)/(x+1)-(2)/(y-1)=(1)/(2);(10)/(x+1)+(2)/(y-1)=(5)/(2),x!=-1,y!=1

Solve the following system of equations: (5)/(x+1)-(2)/(y-1)=(1)/(2),quad (10)/(x+1)+(2)/(y-1)=(5)/(2) where x!=-1 and y!=1

( 5 ) /( x+ 1 ) - (2)/(y - 1 ) = (1)/(2) , (10)/(x + 1) + ( 2 ) /( y - 1) = ( 5)/(2), x ne - 1 and y ne 1

Solve the following system of equations: 5/(x+1)-2/(y-1)=1/2,\ \ \ (10)/(x+1)+2/(y-1)=5/2 , where x!=-1 and y!=1

Solve : (1)/(2x)+(1)/(4y)-(1)/(3z)=(1)/(4),(1)/(x)=(1)/(3y),(1)/(x)-(1)/(5y)+(4)/(z)=2 (2)/(15)

{:((5)/(x - 1) + (1)/(y - 2) = 2),((6)/(x - 1) - (3)/(y - 2) = 1):}

If (5)/(x-1)+(1)/(y-2)=2,(6)/(x-1)+(-3)/(y-2)=1 then x=……….

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

Solve for x and y : (6)/(x-1)-(3)/(y-2)=1 (5)/(x-1)-(1)/(y-2)=2 , where x ne 1, y ne 2 .

The locus of the foot of the perpendicular from the origin on each member of the family (4a+ 3)x - (a+ 1)y -(2a+1)=0 (a) (2x-1)^(2) +4 (y+1)^(2) = 5 (b) (2x-1)^(2) +(y+1)^(2) = 5 (c) (2x+1)^(2)+4(y-1)^(2) = 5 (d) (2x-1)^(2) +4(y-1)^(2) = 5