Question 488231
Let {{{ x }}} = liters of mixture to be drained
and replaced with pure antifreeze
given:
{{{ .4*15 = 6 }}} liters of antifreeze in the system at beginning
{{{ .4x }}} = liters of antifreeze that will be removed by draining
{{{ 6 - .4x }}} = liters of antifreeze left in system after draining
{{{ 6 - .4x + x }}} =  liters of antifreeze left in system after draining
and replacing with pure antifreeze
------------
In words:
(liters of antifreeze after draining & replacing)/(total liters in system after draining) = 50%
{{{ ( 6 + .6x ) / 15 = .5 }}} 
{{{ 6 + .6x = .5*15 }}}
{{{ 6 + .6x = 7.5 }}}
{{{ .6x = 7.5 - 6 }}}
{{{ .6x = 1.5 }}}
{{{ x = 2.5 }}}
2.5 liters need to be drained and replaced with pure antifreeze
check:
{{{ 15 - 2.5 = 12.5 }}} liters of system is 40% antifreeze
{{{ .4*12.5 = 5 }}} liters antifreeze
{{{ 5 + 2.5 = 7.5 }}}
{{{ 7.5 / 15 = .5 }}}
{{{ 7.5 = 7.5 }}}
OK