Question 972698
{{{ .2*18 = 3.6 }}} quarts of antifreeze at beginning
Let {{{ x }}} = quarts of antifreeze solution to be drained off
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Note that you start with {{{ 18 }}} quarts
and you end up with {{{ 18 }}} quarts
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You are draining off a 20% solution, so you are
draining off {{{ .2x }}} quarts of the pure antifreeze
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You replace the solution drained off with {{{ x }}}
quarts of pure antifreeze
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I can now say:
{{{ ( 3.6 - .2x + x ) / 18 = .39 }}}
{{{ ( 3.6 + .8x ) / 18 = .39 }}}
{{{ 3.6 + .8x = .39*18 }}}
{{{ .8x = 7.02 - 3.6 }}}
{{{ .8x = 3.42 }}}
{{{ x = 4.275 }}}
4.275 quarts of solution must be drained off and replaced
with 100% pure antifreeze 
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check:
{{{ ( 3.6 + .8x ) / 18 = .39 }}}
{{{ ( 3.6 + .8*4.275 ) / 18 = .39 }}}
{{{ ( 3.6 + 3.42 ) / 18 = .39 }}}
{{{ 7.02 = .39*18 }}}
{{{ 7.02 = 7.02 }}}
OK