Question 1201248
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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?
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<pre>
Let x be his usual rate, in miles per hour.


The time to travel 18 miles at his usual rate is  {{{18/x}}}  hours.


The time to travel with the faster rate, (x+0.5) miles per hour, is  {{{18/(x+0.5)}}} hours.


The longer time is  {{{1/2}}} of an hour longer:

    {{{18/x}}} - {{{18/(x+0.5)}}} = {{{1/2}}}  of an hour.    (1)


    +----------------------------------------------+
    |    At this point, the setup is complete:     |
    |    you just have an equation to solve.       |
    +----------------------------------------------+


To solve it, multiply both sides by 2x*(x+0.5) = 2x^2 + x.  You will get

    36(x+0.5) - 36x = x^2 + 0.5x

    36x + 18 - 36x = x^2 + 0.5x

    x^2 + 0.5x - 18 = 0

    {{{x[1,2]}}} = {{{(-0.5 +- sqrt(0.5^2 + 4*1*18))/2}}} = {{{(-0.5 +- sqrt(72.25))/2}}} = {{{(-0.5 +- 8.5)/2}}}.


Obviously, only positive root fits  x = {{{(-0.5 + 8.5)/2}}} = {{{8/2}}} = 4  miles per hour.


Check it, by substituting into equation (1)

    {{{18/4 - 18/4.5}}} = 4.5 - 4 = 0.5 = {{{1/2}}}  of an hour.   ! correct !


<U>ANSWER</U>.  The usual rate is 4 miles per hour.
</pre>

Solved.