Should this question read 3/x + 5/y = 9/x - 5/y = 1? If so, here's the solution. [If not, the equation becomes too complicated to be a simultaneous equation at your expected grade.]
3/x + 5/y = 1 and 9/x - 5/y = 1 are the two simultaneous equations. We need to get rid of the fractions so we multiply both sides of both equations by the lowest common denominator, xy.
3y+5x=xy and 9y-5x=xy. By adding these two equations together we get 12y=2xy. We can divide both sides of the equation by 2y, because we know from the original question that y can't be zero (we can't divide by zero). The y's drop out and we get 6=x, therefore x=6. 3(3y+5x)=3xy so 9y+15x=3xy when we multiply the first equation by 3, and that makes two equations starting with 9y, so we can subtract the second from the first giving us 20x=2xy (15x-(-5x) is 15x+5x), so dividing both sides by 2x we get 10=y, therefore y=10. If we substitute these in the original equations we can see that 3/6 + 5/10=1=9/6 - 5/10 (1/2 + 1/2=1 and 3/2 - 1/2=1). So the solution is correct.