Question 816165
{{{ .6*3.6 = 2.16 }}} L antifreeze
originally in radiator
Let {{{ x }}} = liters of solution to
be drained and replaced with water
{{{ .6x }}} = liters of antifreeze 
drained from radiator 
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{{{ ( 2.16 - .6x ) / 3.6 = .5 }}}
Note that you start with 3.6 liters
and end with 3.6 liters
{{{ 2.16 - .6x = .5*3.6 }}}
{{{ .6x = 2.16 - 1.8 }}}
{{{ .6x = .36 }}}
{{{ x = .6 }}}
.6 liters must be drained and replaced with 
pure water
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check:
{{{ .6*.6 = .36 }}} liters of alcohol is drained
There is {{{ 3.6*.6 - .36 }}}
{{{ 2.16 - .36 = 1.8 }}} liters alcohol left
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{{{ .6 }}} liters of water is added to make
{{{ 1.8 }}} liters alcohol and {{{ 3.6 - 1.8 = 1.8 }}}
liters water
They are 50% - 50%
OK