Question 1094059
<br>The trains together cover 288 miles in 3 hours, so their combined speed is
{{{288/3 = 96}}}<br>
The speed of one train is 6 mph slower than the other; so you need two speeds that differ by 6 and add to 96.  If we use f and s for the speeds of the fast and slow trains, then we could use formal algebra and say
{{{f+s = 96}}}
{{{f-s = 6}}}
{{{2f = 102}}}  (add the two equations, to eliminate s)
{{{f = 51}}}
{{{s = 96-f = 45}}}<br>
Once I got the combined speed of96 for the two trains,I would be much less formal than that in solving the problem.  I would simply "take away" the 6 "extra" mph of the faster train, leaving me with two trains going the same speed, and now with a combined speed of 96-6=90 mph.  That would make the slower train going 90/2=45 mph, which means the fast train is going 45+6=51 mph.