Question 659250
Your equation:
{{{ d = 9t[1] }}}
Your friend's equation:
{{{ d = 12t[2] }}}
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{{{ 9t[1] = 12t[2] }}}
{{{ t[2] = (3/4)*t[1] }}}
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This doesn't take into account the extra hour 
which the friend takes, so I add 1 hr to his time
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Time to catch up occurs when
{{{ (3/4)*t[1] + 1 = t[1] }}}
{{{ (1/4)*t[1] = 1 }}}
{{{ t[1] = 4 }}} hrs
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The friend catches up in 4 hrs
check:
{{{ d = 9t[1] }}}
{{{ d = 9*4 }}}
{{{ d = 36 }}} mi
and
{{{ d = 12t[2] }}}
{{{ d = 12*(3/4)*t[1] }}}
{{{ d = 9*t[1] }}}
{{{ d = 9*4 }}}
{{{ d = 36 }}}
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This is done a little differently than the
problem expects, but I think it's right.