bicycles 14mph with no wind. Against the wind, bikes 10 mi inthe same time it takes to bike 20 mi with the wind. What is the speed ofthe wind?​

recall your d = rt, distance = rate * time.w = speed of the wind.14 = speed of the bicycle without wind.now, against the wind, the bicycle is not really going 14 mph fast, is really going "14 - w", since the wind is eroding speed from it, likewise, when the bicycle is going with the wind is not going 14 mph fast either, is really going "14 + w" due to the wind adding speed, let's say it took "t" hours against and also "t" hours with it.[tex]\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{against the wind}&10&14-w&t\\ \textit{with the wind}&20&14+w&t \end{array}~\hfill \begin{cases} 10=(14-w)t\\ \frac{10}{14-w}=\boxed{t}\\ \cline{1-1} 20=(14+w)t \end{cases}[/tex][tex]\bf 20=(14+w)t\implies \cfrac{20}{14+w}=t\implies \stackrel{\textit{substituting in the 2nd equation}}{\cfrac{20}{14+w}=\boxed{\cfrac{10}{14-w}}} \\\\\\ 280-20w=140+10w\implies 280=140+30w\implies 140=30w \\\\\\ \cfrac{140}{30}=w\implies \cfrac{14}{3}=w\implies 4\frac{2}{3}=w[/tex]