Question 203761
<font face="Garamond" size="+2">


Let *[tex \Large S] represent the time, in minutes, it took to finish the second half of the exam.  Then we are given that *[tex \Large \frac{2}{3}\cdot S] is the time, in minutes, it took to finish the first half of the exam.  Add these two quantities for the total time, again in minutes, to finish the exam which is 60 because there are 60 minutes in 1 hour.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}\cdot S + S\ =\ 60]


But *[tex \Large S\ =\ 1\cdot S\ =\ \frac{3}{3}\cdot S], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}\cdot S + \frac{3}{3}\cdot S\ =\  \frac{5}{3}\cdot S\ =\ 60]


So far, so good.


There are two ways to go about solving from here.  You can either multiply both sides by 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3\cdot\frac{5}{3}\cdot S\ =\ 3\cdot60]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5S\ =\ 180]


And then multiply both sides by *[tex \Large \frac{1}{5}]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{5}\cdot 5S\ =\ \frac{1}{5}\cdot 180]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S\ = \ 36]


Or, you can combine those two steps into one, as your book shows, but multiplying by the reciprocal of the fractional coefficient on the variable. (That is because we know that *[tex \Large a\cdot\frac{1}{a}\ =\ 1\ \ \forall\ a\ \in\ \R])


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{3}{5}\cdot\frac{5}{3}\cdot S\ =\ \frac{3}{5}\cdot 60]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S\ = \ 36]


Achieving the same result, as one would expect.



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