" 18."(y-1)/(3)-(y-2)/(4)=1

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

If 4x+3y=18xy and 2x-5y+4xy=0 then the values of x and y will be respectively -(1)/(2) and -(1)/(3)(b)-1 and -3(c)(1)/(2) and (1)/(3) (d) (1)/(4) and (1)/(3)

1/(3x+y)+1/(3x-y)=3/4 and 1/(3x+y)-1/(3x-y)= -1/4

(1)/(2(3x+4y))=(1)/(5(2x-3y))=(1)/(4),

Simplify: (y^3-2y^2+3y-4) (y-1) - (24-3)(y^2-y+1)