Q. 361 - Intermediate Algebra

Question

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 in2 hoursinto a headwind.Find the speed of the jet in still air and the speed of the wind

Step-by-Step Solution

Answer

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

Step 1.Given information

Let x represents the speed of the jet in still air.

Let y represents the speed of the wind.

The following chat will help us organize the data.

The jet makes two trips.

One is tailwind and one is headwind.

In a tailwind,the wind helps the jet so the rate is $x+y$.

And in a headwind,the wind slows the jet and so  the rate is $x-y$.

Each trip makes 2 hours.

In a tailwind the jet flies 804 miles.

In a headwind the jet flies 776 miles.

Name	Rate	Time	Distance D=r.t
Tailwind	x+y	2	$$\begin{gathered} 2(x+y)=804 \\ (x+y)=\frac{804}{2} \\ (x+y)=402 \end{gathered}$$
Headwind	x-y	2	$$\begin{gathered} 2(x-y)=776 \\ (x-y)=\frac{776}{2} \\ x-y=388 \end{gathered}$$

 

Step 2.Solve for x.

From the equations

$$\begin{gathered} x+y=402 \\ x-y=388 \\ ----- \\ 2x=709 \\ x=395 \end{gathered}$$

Step 3.Solve for y.

Substitute $x=395$ into the equation

$$\begin{gathered} x+y=402 \\ 395+y=402 \\ y=402-395 \\ y=7 \end{gathered}$$

Step 4.Check

Substitute $x=395,y=7$ into the equation $x+y=402$.

$$\begin{gathered} x+y=402 \\ 395+7=402 \\ 402=402 \end{gathered}$$

This is true.

Substitute $x=395,y=7$ into the equation $x-y=388$.

$$\begin{gathered} x-y=388 \\ 395-7=388 \\ 388=388 \end{gathered}$$

this is true.

Therefore,the speed of the jet still in air is 395 mph and the speed of the wind is 7 mph.