Question 898678
Let {{{ d }}} = distance faster car travels in miles
{{{ d - 30 }}} = distance slower car travels in miles
Let {{{ t }}} = time in hours for the cars to be 30 mi apart
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Faster car:
{{{ d = 34t }}}
Slower car:
{{{ d - 30 = 24t }}}
{{{ d = 24t + 30 }}}
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By substitution:
{{{ 34t = 24t + 30 }}}
{{{ 10t = 30 }}}
{{{ t = 3 }}}
They will be 30 mi apart in 3 hrs
check:
{{{ d = 34t }}}
{{{ d = 34*3 }}}
{{{ d = 102 }}} mi
and
{{{ d = 24t + 30 }}}
{{{ d = 24*3 + 30 }}}
{{{ d = 72 + 30 }}}
{{{ d = 102 }}} mi