An algebra math question that i need help on

an automobile radiator contains 16 litres of antifreeze and water. this mixture is 30% antifreeze. how much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze

Let D liters be the amount drained
Antifreeze drained = 0.3D liters
Antifreeze gained after re-filling = D liters
Original amount of anti-freeze = 0.3*16 = 4.8 liters
Final amount required = 8 liters
8-4.8 = D - 0.3D = 0.7D
D = 3.2/0.7 = 4.57 liters

well answer like this: (.30/16-.30/x)+1/x=.50/16

1.6 litres??