In the following exercises, translate to a system of equations and solve

A passenger jet can fly 804 miles in 2 hours with a tailwind but only 776 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

According to the statements of the question we have to find out he speed of the jet in still air and the speed of the wind.

Let us take j=speed of wind in still sir

w= speed of the wind

In a tailwind, the rate is j + w.

In a headwind, the rate is j − w and each trip takes 2 hours .

So we can write $$\begin{gathered} 2(j+w)=804\sin cedis\tan ce=rate\times time \\ 2(j-w)=776 \end{gathered}$$

So this is our system of equations.

The system of equations are

$$\begin{gathered} 2(\mathrm{j}+\mathrm{w})=804\Rightarrow 2\mathrm{j}+2\mathrm{w}=804 \\ 2(\mathrm{j}-\mathrm{w})=776\Rightarrow 2\mathrm{j}-2\mathrm{w}=776 \end{gathered}$$

When we add both equations we get value for j.

At the tailwind, the actual rate of the jet is 395 + 7 = 402 mph. In 2 hours the jet would travel$402\times 2=804miles$.

In a headwind, the jet’s actual rate is 395 − 7 = 388 mph. In 2hours the jet would travel $388\times 2=776miles$

Hence the rate of the jet in still air is 395mph and the speed of wind is 7mph.