Question 319077
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Let *[tex \Large r] represent the rate for the first part of the trip.  Then *[tex \Large r\ -\ 5] represents the rate on the second part of the trip.  Let *[tex \Large t] represent the time elapsed on the first part of the trip, then *[tex \Large 3\ -\ t] represents the time elapsed on the second part of the trip.

The first part of the trip is described by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 69\ =\ rt]

Which can be more conveniently expressed as:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{69}{r}]

The second part of the trip is described by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 23\ =\ (r\ -\ 5)(3\ -\ t)]

and which can more conveniently be expressed as:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\ -\ t\ =\ \frac{23}{r\ -\ 5}]

but which requires a bit more manipulation so that we can develop an expression equal to *[tex \Large t]:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 3\ -\ \frac{23}{r\ -\ 5}]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{3(r\ -\ 5)\ -\ 23}{r\ -\ 5}]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{3r\ -\ 38}{r\ -\ 5}]

Now that we have two expressions equal to *[tex \Large t] we can set them equal to each other:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{69}{r}\ =\ \frac{3r\ -\ 38}{r\ -\ 5}]

Cross multiply and put the quadratic into standard form:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 69r\ -\ 345\ =\ 3r^2\ -\ 38r]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r^2\ -\ 107r\ +\ 345\ =\ 0]

This one requires the quadratic formula.  There are a number of on-line solvers but the one I like is:  http://zonalandeducation.com/mmts/miscellaneousMath/quadraticRealSolver/quadraticRealSolver.html

Since the determinant is prime, and you were asked for the answer to the nearest hundredth, just use the numerical approximations presented. Note that there are two roots.  However, remember that r was defined to be the faster of the two rates, and the small value root is less than 5.  That means when you subtract 5 from that value, the slower speed would actually be backwards.  Discard the smaller root in favor of the larger.

By the way, rounding this answer to the nearest 100th is pure silliness.  As a general rule, you should <i>never</i> express the result of a calculation involving measurement to greater precision than the <i>least</i> precise given/empirical measurement.  None of your given data was more precise than the nearest whole number, hence the rate should have been expressed to the nearest whole number at best.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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