Take a message soldier
A messenger runs from the rear to the head of a marching column and
back. When he gets back, the rear is where the head was when he set
off. What is the ratio of his speed to that of the column?
Problem
Image

During an exercise a column of soldiers were marching at a
constant speed. At the end of the column was a messenger. He was
commanded to proceed to the head of the column, deliver a dispatch
and return to his former position at the back of the column,
without delay.
On regaining his former position he noticed that the end of
the column was in the same position that the head of the column had
occupied when he first left his own position.
What is the ratio of the messenger's speed to that of the
column whilst this order was being carried out? (Assume that no
time was wasted in delivering the message or returning to his
former position.)

Getting Started
Find the time taken for the messenger to march up to the head of the column, then add the time for him to march back. This is equal to the time taken by the column to advance a distance equal to the length of the column.
Student Solutions
First of all we need to assign letters to the important variables in the problem.
Let $v$ be the speed of the column.
Let $u$ be the speed of the messenger.
Let $L$ be the length of the column.
We can assume that the messenger travelled at a constant speed because he was ordered to deliver the message and return to his former position "without delay". Assume that the speed $u$ is the messenger's maximum speed which is constant (he does not get tired, for instance). We need to find ${u\over v}$.
When delivering the message, the messenger travels at a speed $uv$ relative to the column. He travels a distance $L$ relative to the column. Therefore, the time to reach the head of column is $${L\over uv}\;.$$
When returning to his position, the messenger travels at a speed $u+v$ relative to the column. Again, he travels a distance $L$ relative to the column. Therefore, the time to return to the end of column is $${L\over u+v}\;.$$
While the messenger is carrying out his mission, the end of the column travels a distance $L$.
The time for the column to travel this distance is $${L \over v}\;.$$
The time for the messenger to complete the mission is equal to the time for the column to travel the distance $L$.
$$\begin{eqnarray} {L \over uv} + {L \over u+v} &=& {L\over v}\\ \Rightarrow {1\over uv} + {1\over u+v} &=& {1\over v}\qquad\, \mathrm{[divide\ by\ L]}\\ \Rightarrow {v\over uv} + {v\over u+v} &=& 1\qquad\;\; \mathrm{[multiply\ by\ v]}\\ \end{eqnarray}$$
As stated above, we need to find ${u\over v}$. If we divide top and bottom of the fractions on the left hand side by $v$ and then substitute $x$ for ${u\over v}$, we get the following equation.
$$\begin{align} {1\over x1} + {1\over x+1} &= 1\\ \Rightarrow (x+1) + (x1) &= (x+1)(x1)\qquad \mathrm{[multiply\ through\ by\ (x+1)(x1)]}\\ \Rightarrow 2x &= x^2  1\qquad\qquad\quad\; \mathrm{[expand\ the\ brackets\ and\ gather\ terms]}\\ \Rightarrow 0 &= x^2  2x  1 \end{align}$$
We now have a quadratic equation, for which we can use the standard formula to solve.
$$\begin{align} x &= \frac{2 \pm \sqrt{4 + 4}}{2}\\ x &= \frac{2 \pm \sqrt{8}}{2}\\ x &= 1 \pm \sqrt{2} \end{align}$$
as $\sqrt{8}$ is equal to $2\sqrt{2}$.
We now have to consider the two possible solutions. The ratio cannot be negative, because the velocity of the messenger is in the same direction as the velocity of the column. This means that the ratio cannot be $1  \sqrt{2}$. Thus, the ratio of the messenger's speed to that of the column is $x=1 + \sqrt{2}.$
Teachers' Resources
Why do this problem?
An exercise in relative velocity and use of the distance time formula. Learners have to read and make sense of the question and interpret the information in an equation.Key questions
What are the unknowns? What notation shall we use?What are we asked to find out? Which unknown(s) does that involve.
What does the question tell us about distance, velocity and time?
Can we write an equation from the information given?
Are all the solutions of the equation valid? Why/why not?