Question 786612
Let {{{ x }}} = liters of coolant ( water + antifreeze ) to
be drained and replaced with pure water
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{{{ .7*4.8 = 3.36 }}} liters of antifreeze originally in radiator
{{{ .7x }}} = liters of antifreeze in {{{ x }}} liters of coolant that was drained out
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What helps in this problem is that you start with 4.8 liters in radiator,
and you end with 4.8 liters
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{{{ ( 3.36 - .7x ) / 4.8 = .5 }}}
{{{ 3.36 - .7x = .5*4.8 }}}
{{{ 3.36 - .7x = 2.4 }}}
{{{ .7x = 3.36 - 2.4 }}}
{{{ .7x = .96 }}}
{{{ x = 1.371 }}}
1.371 liters must be drained out and replaced with water
check:
{{{ ( 3.36 - .7x ) / 4.8 = .5 }}}
{{{ ( 3.36 - .7*1.371 ) / 4.8 = .5 }}}
{{{ ( 3.36 - .96 ) / 4.8 = .5 }}}
{{{ 2.4 / 4.8 = .5 }}}
{{{ 2.4 = 2.4 }}}
OK