Question 982020
Variables:
r = rate with no wind
t = time it takes to cycle 48 miles with the wind, or 30 miles against the wind.


Table:


<table border=1><tr><th></th><th>With Wind</th><th>Against Wind</th></tr><tr><td>Speed</td><td>r+3</td><td>r-3</td></tr><tr><td>Distance</td><td>48</td><td>30</td></tr><tr><td>Time</td><td>t</td><td>t</td></tr></table>


The table isn't mandatory, but it helps sort out all the data (known or unknown).


Focus on the "With Wind" column. This is where the wind helps and pushes him faster. His speed increases from r to r+3. He can go d = 48 miles. The time taken is t.


Use {{{d = r*t}}} and the info above and solve for t


{{{d = r*t}}}


{{{48 = (r+3)*t}}}


{{{48/(r+3) = ((r+3)*t)/(r+3)}}}


{{{48/(r+3) = t}}}


{{{t = 48/(r+3)}}}


The time expression to go 48 mi with the wind is {{{48/(r+3)}}} hours.

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Now move to the next column "against wind"


The wind is now slowing him down. His speed decreases from r to r-3. He can go d = 30 miles against the wind. The time taken is t because it takes the same amount of time to go 48 mi with the wind and 30 mi against the wind.


Use {{{d = r*t}}} and the info above and solve for t



{{{d = r*t}}}


{{{30 = (r-3)*t}}}


{{{30/(r-3) = ((r+3)*t)/(r+3)}}}


{{{30/(r-3) = t}}}


{{{t = 30/(r-3)}}}



The time expression to go 30 mi against the wind is {{{30/(r-3)}}} hours.

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From the two sub-parts above, we found these two equations
{{{t = 48/(r+3)}}} and {{{t = 30/(r-3)}}}


Because the t values are the same, we can equate the two right hand sides to get {{{48/(r+3) = 30/(r-3)}}}. It doesn't matter which side what expression is on.


Now we solve for r. There are a few ways to do this. One way is to cross multiply.



{{{48/(r+3) = 30/(r-3)}}}


{{{48(r-3) = 30(r+3)}}} Cross multiplication happens here


{{{48r-144 = 30r+90}}}


{{{48r-144-30r = 30r+90-30r}}}


{{{18r-144 = 90}}}


{{{18r-144+144 = 90+144}}}


{{{18r = 234}}}


{{{18r/18 = 234/18}}}


{{{r = 234/18}}}


{{{r = 13}}}



Question: What is his normal bicycling speed with no wind?
Answer: <font size=5 color="red">13 mph</font>


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Extra:


If you wanted to find the time (t), you would pick on any equation that has r and t in it. Plug in r = 13 and evaluate.



{{{t = 48/(r+3)}}}


{{{t = 48/(13+3)}}}


{{{t = 48/(16)}}}


{{{t = 4}}}



It takes him 4 hours to either go 48 mi with the wind and 30 mi against the wind. The choice of equation doesn't matter because this (r,t) pair makes both equations true at the same time. They both have the same r and t values.