[(5)/(x+1)y=1],[(10)/(x+1)+(2)/(y-1)=(5)/(2)" whose "x!=-1" and "y!=0]

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

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

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

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

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

If 5^(x)=(0.5)^(y)=1000,then(1)/(x)-(1)/(y)=

If ((x_(1),x_(2)),((y_(1),y_(2))-((2,3),(0,1)),=((3,5),(1,2)) then find x_(1),x_(2),y_(1),y_(2) ,

The equation of line whose midpoint (x_(1),y_(1)) is in between the axes, is.............. A) (x)/(x_(1))+ (y)/(y_(1)) =2 B) (x)/(x_(1))+ (y)/(y_(2)) =1/2 C) (x)/(x_(1))+ (y)/(y_(2)) =1/2 D) None of these