Question 1009374
Let the rate of the slower car be r.  The faster moves at r+8.
In general rt = d. Thus t = d/r.  Here the times are equal, so we write
{{{(r+8)/238 = r/210}}}
We cross-multiply and solve giving us
210(r+8) = 238r
210r + 1680 = 238r
28r = 1680
r = 60 mph
r+8 = 68 mph