Question

Against the wind a commercial airline in South America flew 504 miles in 3.5 hours. With a tailwind the return trip took 3 hours. What was the speed of the plane in still​ air? What was the speed of the​ wind?

Answer 1

Recall your d = rt, distance = rate * time

so, notice, it went 504 against the wind, and the return trip, well, unless the road elongated or shrunk somewheres, the return trip is also 504 miles.

now, if say the plane has a still air speed of "p", and the wind has a speed of "w", when it was going against the wind, it wasn't really going "p" fast, it was going "p - w" fast, because the wind was subtracting speed from it.

now, when it was going with the wind, it wasn't going "p" fast either, it was going faster, at "p + w" fast, because the wind was adding to it. thus

$\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{against the wind}&504&p-w&3.5\\ \textit{with the wind}&504&p+w&3 \end{array} \\\\\\ \begin{cases} 504=3.5(p-w)\implies \frac{504}{3.5}=p-w\\ \qquad 144=p-w\implies \boxed{w}=p-144\\\\ 504=3(p+w)\\ ----------\\ \frac{504}{3}=p+w\implies 168=p+w\\ \qquad 168=p+\boxed{p-144} \end{cases} \\\\\\ 168=2p-144\implies \cfrac{168+144}{2}=p\implies 156=p$