Question 225271
Let {{{x}}} = liters of  mixture to be drained 
and replaced with pure antifreeze.
In words:
(liters of antifreeze I end up with)/(final total liters in radiator) = 50%
Notice that I start with 16 liters in the radiator
and I end up with 16 liters in the radiator, so I can write
(liters of antifreeze I end up with)/16 = 50%
16 liters of 30% antifreeze has {{{.3*16 = 4.8}}} liters antifreeze
I drain out {{{x}}} liters, and in doing so, I remove .3x liters
antifreeze
So far, I've got {{{4.8 - .3x}}} liters antifreeze in radiator
Now I pour in {{{x}}} liters of antifreeze
I end up with {{{4.8 - .3x + x}}} liters antifreeze
My equation is then
{{{(4.8 + .7x)/16 = .5}}}
{{{4.8 + .7x = 8}}}
{{{.7x = 3.2}}}
{{{x = 4.571}}} liters
4.571 liters need to be drained out and
replaced with pure antifreeze
check answer
If I replace {{{x}}} with a 30% mixture again, I'll be back
to 30% in the whole radiator
{{{.3*4.571 = 1.371}}} liters antifreeze
{{{16 - 4.571 = 11.429}}}
{{{.3*11.429 = 3.429}}} liters antifreeze in rest of radiator 
{{{(1.371 + 3.429)/16 = .3}}}
{{{4.8/16 = .3}}}
{{{4.8 = 4.8}}}
OK