Question 1201248
<pre>
Hello I'm having trouble setting this problem up. Any help would be appreciated. Thanks.
 A traveler having 18 miles to go, calculates that his usual rate would make
him one-half hour late for an appointment; he finds that in order to arrive on
time he must travel at a rate one-half mile an hour faster. What is his usual
rate?

Let usual/regular speed be S
Then time taken to get to appointment = {{{18/S}}}
Increasing speed by {{{matrix(1,2, 1/2, mph)}}}, or .5 mph means that his new speed to get there on-time is: (S + .5) mph.
So, time he'd take to get to the appointment, based on his new speed = {{{18/(S + .5)}}}
At the new speed , he'll get there on-time by "shaving" {{{1/2-hour}}}, or .5 hour off of the previous late-time, or {{{18/S - .5}}}
We then get the following TIME equation: {{{matrix(1,3, 18/(S + .5), "=", 18/S - .5)}}}
                                          {{{matrix(1,3, 18(S), "=", 18(S + .5) - .5S(S + .5))}}} ------ Multiplying by LCD, {{{S(S + .5)}}}
                                  {{{matrix(3,3, 18S, "=", 18S + 9 - .5S^2 - .25S, 0, "=", 9 - .5S^2 - .25S, .5S^2 + .25S - 9, "=", 0)}}}  
                                     {{{matrix(1,3, 2S^2 + S - 36, "=", 0)}}} ----- Multiplying by 4
                                 (S - 4)(2S + 9) = 0 ----- Factoring trinomial
                                  S - 4 = 0            or      2S + 9 = 0
           <font size = 4><font color = red><b>Normal/Usual speed</font></font></b>, or S = <font size = 4><font color = red><b>4 mph</font></font></b>       or      2S = - 9 (ignore)</pre>