Question 1167191
<br>
Let the speed going be x and the speed going be x-15.<br>
The time going is then 140/x and the time returning is 140/(x-15).<br>
The time returning is 3 hours more than the time going:<br>
{{{140/(x-15) = 140/x+3}}}<br>
{{{140/(x-15) = (140+3x)/x}}}<br>
{{{140x = (140+3x)(x-15)}}}
{{{140x = 3x^2+95x-2100}}}
{{{3x^2-45x-2100 = 0}}}
{{{x^2-15x-700 = 0}}}<br>
{{{(x-35)(x+20) = 0}}}<br>
{{{x = 35}}} or {{{x = -20}}}<br>
Clearly the negative solution makes no sense in the problem....<br>
ANSWERS:
speed going = x = 35mph
speed returning = x-15 = 20mph<br>
CHECK:
time going = 140/35 = 4 hours
time returning = 140/20 = 7 hours = 3 hours more than time going<br>
That's a formal algebraic solution, which involves a lot of work -- especially factoring the quadratic into two linear factors.<br>
If a formal algebraic solution is not required, a little mental arithmetic will find the answer much faster and with much less effort.  Simply look for combinations of hours and miles per hour that make 140 miles.<br>
Two easy combinations that satisfy all the conditions of the problem are
(1) 4*35 = 140 --> 4 hours at 35mph
(2) 7*20 = 140 --> 7 hours at 20mph<br>