Question 213393
the best way (for me) to solve this is to
express an equation in words that sum up the 
whole situation.
The 1st thing to realize is that you start out with
6 liters of fluid and you end up with 6 liters
of fluid
Here it is in words:
((original amount of antifreeze) - (amount of antifreeze drained out) + (amount of antifreeze put back in)) 
[all divided by]
(total amount of fluid I end up with) = 90%
-------------------------------------------
Let {{{x}}}= liters of fluid I drain out and replace with pure antifreeze
Original amount of antifreeze = {{{.8*6 = 4.8}}} liters
Amount of antfreeze drained out = {{{.8x}}}
Amount of antifreeze put back in = {{{x}}}
------------------------------------------
Now I can write the equation
{{{(4.8 - .8x + x)/6 = .9}}}
{{{4.8 - .8x + x = 5.4}}}
{{{.2x = 5.4 - 4.8}}}
{{{.2x = .6}}}
{{{x = 3}}}
3 liters, or 1/2 the fluid in the radiator must
be drained and replaced with pure antifreeze to get
90% antifreeze
check answer:
There must be 10% water in the final solution
I start with {{{.2*6 = 1.2}}} liters water
I drain out {{{.2x = .2*3}}} or {{{.6}}} liters water
I end up with {{{1.2 - .6 = .6}}} liters water
{{{.6 = .1*6}}}
So, I do end up with 10% water