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Sagot :
We are given that there are 4 cities that lie on a straight highway. We can visulaize our problem to be 1-D line. We will go ahead and place these cities on a line as follows:
The above placement of 4 cities is NOT to scale. We will go ahead and update our graphics with each step.
[tex]\text{\textcolor{#FF7968}{Distance from City A to City B is 5 times the distance from City B to City C}}[/tex]We will first go ahead and decrypt the above statement and write it down as a mathematical equation as follows:
[tex]\textcolor{#FF7968}{|AB|}\text{\textcolor{#FF7968}{ = 5}}\textcolor{#FF7968}{\cdot|BC|\ldots}\text{\textcolor{#FF7968}{ Eq1}}[/tex]Where,
[tex]\begin{gathered} |AB|\colon\text{ Distance from City A to B} \\ |BC|\text{ : Distance from City B to C} \end{gathered}[/tex]The next statement is as follows:
[tex]\text{\textcolor{#FF7968}{Distance from City A to City D is 2 times the distance from City A to City B}}[/tex]We will first go ahead and decrypt the above statement and write it down as a mathematical equation as follows:
[tex]\textcolor{#FF7968}{|}\text{\textcolor{#FF7968}{AD| = 2}}\textcolor{#FF7968}{\cdot|AB|\ldots}\text{\textcolor{#FF7968}{ Eq2}}[/tex]Where,
[tex]|AD|\colon\text{ Distance from City A to City D}[/tex]The last statement states as follows:
[tex]\text{\textcolor{#FF7968}{The distance from City C to City D is 16 miles}}[/tex]We will first go ahead and decrypt the above statement and write it down as a mathematical equation as follows:
[tex]\textcolor{#FF7968}{|CD|}\text{\textcolor{#FF7968}{ = 16 }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Eq3}}[/tex]Where,
[tex]|CD|\text{ : Distance from City C to City D}[/tex]WE have the following three equations summarized as follows:
[tex]\begin{gathered} |AB|\text{ = 5}\cdot|BC| \\ |AD|\text{ = 2}\cdot|AB| \\ |CD|\text{ = 16} \end{gathered}[/tex]We will now update our graphical representation.
Substitute the (Eq1) and (Eq2) as follows:
[tex]\begin{gathered} |\text{ AD | = 2}\cdot(\text{ 5}\cdot|BC|\text{ ) } \\ \textcolor{#FF7968}{|AD|}\text{\textcolor{#FF7968}{ = 10}}\textcolor{#FF7968}{\cdot|BC|} \end{gathered}[/tex]The distance ( AD ) can be split into three parts as follows:
[tex]\begin{gathered} |\text{ AD | = | AB | + | BC | + |CD|} \\ \text{Then use the above manipulation,} \\ \text{| AB | + | BC | + |CD| = 10}\cdot\text{ ( }|BC|\text{ )} \\ \textcolor{#FF7968}{|AB|}\text{\textcolor{#FF7968}{ = 9}}\textcolor{#FF7968}{\cdot|BC|}\text{\textcolor{#FF7968}{ - |CD|}} \end{gathered}[/tex]Substitute Eq1 and Eq3 into the above result:
[tex]\begin{gathered} 5\cdot|BC|\text{ = 9}\cdot|BC|\text{ - 16} \\ 4\cdot|BC|\text{ = 16} \\ \textcolor{#FF7968}{|BC|}\text{\textcolor{#FF7968}{ = 4 miles}} \end{gathered}[/tex]Use Eq1 to evaluate ( AB ) as follows:
[tex]\begin{gathered} |AB|\text{ = 5}\cdot|BC| \\ |AB|\text{ = 5}\cdot4 \\ \textcolor{#FF7968}{|AB|}\text{\textcolor{#FF7968}{ = 20 miles}} \end{gathered}[/tex]The correspnding inter-city distances are as follows:
[tex]\textcolor{#FF7968}{|AB|}\text{\textcolor{#FF7968}{ = 20 miles, | BC| = 4 miles, |CD| = 16 miles}}[/tex]
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