Respuesta :
recall your d = rt, distance = rate * time
so hmm, if say the speed rate of the wind is "w", when the plane is flying with the wind, is not really flying 325 mph, is really flying " 325 + w " fast.
now, when the plane is going against the wind, so-called upwind, is not really going 325 mph either, is really going " 325 - w " fast.
now, bear also in mind that, it took "t" hours to go one way, and it also took "t" hours to go the other way.
so.... let's check then
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{with the wind}&190&325+w&t\\ \textit{against the wind}&135&325-w&t \end{array} \\\\\\ \begin{cases} 190=t(325+w)\implies \frac{190}{325+w}=\boxed{t}\\\end{cases}[/tex]
[tex]\bf \cfrac{135}{325-w}=\cfrac{190}{325+w}\implies 135(325+w)=190(325-w) \\\\\\ (135\cdot 325)+135w=(190\cdot 325)-190w \\\\\\ 135w+190w=(190\cdot 325)-(135\cdot 325)\impliedby \textit{some common factor} \\\\\\ 325w=325(190-135)\implies w=55[/tex]
so hmm, if say the speed rate of the wind is "w", when the plane is flying with the wind, is not really flying 325 mph, is really flying " 325 + w " fast.
now, when the plane is going against the wind, so-called upwind, is not really going 325 mph either, is really going " 325 - w " fast.
now, bear also in mind that, it took "t" hours to go one way, and it also took "t" hours to go the other way.
so.... let's check then
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{with the wind}&190&325+w&t\\ \textit{against the wind}&135&325-w&t \end{array} \\\\\\ \begin{cases} 190=t(325+w)\implies \frac{190}{325+w}=\boxed{t}\\\end{cases}[/tex]
[tex]\bf \cfrac{135}{325-w}=\cfrac{190}{325+w}\implies 135(325+w)=190(325-w) \\\\\\ (135\cdot 325)+135w=(190\cdot 325)-190w \\\\\\ 135w+190w=(190\cdot 325)-(135\cdot 325)\impliedby \textit{some common factor} \\\\\\ 325w=325(190-135)\implies w=55[/tex]