Question 356982
{{{z/(z+3)=3z/(5z-1)}}}.  Assuming in the emantime that z are not equal to -3 or 1/5, we get {{{5z^2-z = 3z^2+9z}}}.
 {{{5z^2-3z^2-z-9z=0}}},
{{{2z^2 - 10z=0}}},
{{{z^2 - 5z = 0}}},
{{{z(z-5) = 0}}},
z=0 or z=5.  Since neither of z=0 or z=5 are equal to -3 or 1/5, the solution set is {0,5}.