[x+y=5xy],[3x+2y=13xyquad (x!=0,y!=0)]

Solve the following system of equations: x+y=5xy,quad 3x+2y=13xy,quad x!=0,quad y!=0

x + y = 5 xy , 3x + 2y = 13 xy ( x ne 0, y ne 0 )

4 x + 6y = 3 xy , 8x + 9 y = 5xy ( x ne 0 , y ne 0 )

Solve: 3 (2x+ y ) = 7xy 3(x+ 3y ) = 11xy , x ne 0 , y ne 0

if y + x = 5xy , y - x = -xy

In the equations 3x+2y=13xy and 4x-5y=2xy, the values of x and y that satisfy the equations are

Solve for x and y : 2(3x-y)=5xy, 2(x+3y)=5xy ,

Solve for x and y, by reducing the following equations in a pair of linear equations : 2x + 3y = 5xy and 3x - 2y = xy.

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

xy^(2)-x^(2)y-5=0