Question 913981
Let {{{ x }}} = liters of coolant to be drained
{{{ .7x }}} = liters of antifreeze in {{{ x }}} liters
{{{ .7*3.6 = 2.52 }}} = liters of antifreeze 
originally in radiator
---------------
Note that {{{ x }}} liters drained off are
replaced with {{{ x }}} liters of water, so
radiator ends up with {{{ 3.6 }}} liters
---------------
{{{ ( 2.52 - .7x ) / 3.6 = .5 }}}
{{{ 2.52 - .7x = .5*3.6 }}}
{{{ 2.52 - .7x = 1.8 }}}
{{{ .7x = 2.52 - 1.8 }}}
{{{ .7x = .72 }}}
{{{ x = 1.029 }}}
1.029 liters of coolant must be drained
off and replaces with water
--------------------------
check:
{{{ .7*1.029 = .72 }}} liters of coolant drained off
{{{ .3*1.029 = .3087 }}} liters of water drained off
--------------------
{{{ .3*3.6 = 1.08 liters of water to start with
{{{ 1.08 - .3087 = .7713 }}} liters of water remaining
{{{ .7713 + 1.029 = 1.8 }}} liters of water after 
re-filling with pure water
{{{ 3.6 - 1.8 = 1.8 }}} liters of antifreeze left
{{{ 1.8/3.6 = .5 }}}
OK