Question 29351
d = distance from cs to pp
6 mi/hr = average rate cs to pp
15 mi/hr = average rate pp to cs
t + 1.5 = time in hours cs to pp
t = time in hours pp to cs
[1] {{{d = r[1] * t[1]}}}
[2] {{{d = r[2] * t[2]}}}
set  [1] equal to [2]
{{{6 * (t + 1.5) = 15 * t}}}
{{{6 * t + 9 = 15 * t}}}
{{{9 = 9 * t}}}
{{{t = 1}}}
t + 1.5 = time in hours cs to pp
1 + 1.5 = 2.5 hrs 
check
{{{6 * 2.5 = 15 * 1}}}
{{{15 = 15}}}
OK