Question

Flying against the wind, a jet travels 8730mi in 9 hours. Flying with the wind, the same jet travels 7260mi in 6 hours. What is the rate of the jet in still air and what is the rate of the wind?

Rate of the jet in still air: mi/h

Rate of the wind: mi/h

Answer 1

recall your d = rt, distance = rate * time.

j = jet's rate

w = wind's rate

so with the wind the actual speed of the Jet is really j + w, because the wind is adding speed to it, and against it is j - w, since the wind is eroding speed from it.

`\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{against the wind}&8730&j-w&9\\ \textit{with the wind}&7260&j+w&6 \end{array}~\hfill \begin{cases} 8730=(j-w)(9)\\ 7260=(j+w)(6) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{\cfrac{8730}{9}=j-w\implies }970=j-w\implies 970+w=j \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{using the 2nd equation}}{\cfrac{7260}{6}=j+w}\implies 1210=j+w\implies \stackrel{\textit{doing some substitution on \underline{j}}}{1210=(970+w)+w} \\\\\\ 1210=970+2w\implies 240=2w\implies \cfrac{240}{2}=w\implies \boxed{120=w} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{we know that}}{970+w=j}\implies 970+120=j\implies \boxed{1090=j}`