We have to write a problem using rate, time and distance as the context that can be solved with the equation given.
We can start by relating rate (speed), time and distance:
[tex]v=\frac{d}{t}[/tex]this equation represents the average speed, that can be calculated as the quotient between the distance and time.
We can also write, derived from the previous equation:
[tex]\begin{gathered} t=\frac{d}{v} \\ d=v\cdot t \end{gathered}[/tex]The equation t = d/v can be useful in this case, as we can let x represent distance, 1/2 represent total time and 6 and 4 represent speed in different parts of a path.
We then can write something like:
[tex]t=t_1+t_2=\frac{d_1}{v_1}+\frac{d_2}{v_2}[/tex]For example a trip between places A and C. There is a place between A and C, called B.
We walk from A to B at 6 km/h and then from B to C at 4 km/h. It takes half an hour (1/2 hour) to get from A to C.
We also know that the distance from B to C is one kilometer less than the distance from
A to B.
The question is: what is the distance from A to B?
Let x be the distance from A to B.
We can write this problem as:
[tex]\begin{gathered} t=t_{AB}+t_{BC}=\frac{d_{AB}}{v_{AB}}+\frac{d_{BC}}{v_{BC}}_{} \\ \frac{1}{2}=\frac{d_{AB}}{6}+\frac{d_{BC}}{4} \end{gathered}[/tex]As the distance from B to C is one km less than from A to B we can write:
[tex]\begin{gathered} d_{AB}=x \\ d_{BC}=d_{AB}-1=x-1 \end{gathered}[/tex]replacing in the equation, we get:
[tex]\frac{1}{2}=\frac{x}{6}+\frac{x-1}{4}[/tex]that is equivalent to the equation given.
Answer:
The problem can be stated as:
"We have to go from A to C. There is a place between A and C, called B.
We walk from A to B at 6 km/h and then from B to C at 4 km/h. It takes half an hour (1/2 hour) to get from A to C.
We also know that the distance from B to C is one kilometer less than the distance from A to B.
The question is: what is the distance from A to B?"
