Question

The speed of a plane in still air is x miles per hour. The wind blows at a constant speed of y miles per hour in an unvarying direction. Flying into a headwind (against the wind), the plane takes 4 hours to travel 500 miles. Flying with a tailwind (with the wind), the plane takes 3 hours to travel 465 miles. Write a system of linear equations to represent the scenario.(1 point)
Responses

{4(x−y)=5003(x+y)=465
enlarged left brace start layout first row 4 left parenthesis x minus y right parenthesis equals 500 second row 3 left parenthesis x plus y right parenthesis equals 465 end layout

{4x−y=5003x+y=465
enlarged left brace start layout first row 4 x minus y equals 500 second row 3 x plus y equals 465 end layout

{4(x+y)=5003(x−y)=465
enlarged left brace start layout first row 4 left parenthesis x plus y right parenthesis equals 500 second row 3 left parenthesis x minus y right parenthesis equals 465 end layout

{4x+y=5003x−y=465
enlarged left brace start layout first row 4 x plus y equals 500 second row 3 x minus y equals 465 end layout
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option a b c or d

Answer 1

To set up the system of linear equations, we need to analyze the two scenarios given:

1. **Flying into a headwind (against the wind)**: The effective speed of the plane is \( x - y \).

The distance traveled is 500 miles, and the time taken is 4 hours. Therefore, we can use the formula distance = speed × time to write the equation:
\[
4(x - y) = 500
\]

2. **Flying with a tailwind (with the wind)**: The effective speed of the plane is \( x + y \).

The distance traveled is 465 miles, and the time taken is 3 hours. Similarly, we write the equation:
\[
3(x + y) = 465
\]

Now, let's put the equations together:

1. \( 4(x - y) = 500 \)
2. \( 3(x + y) = 465 \)

Thus, the correct system of linear equations is:
\[
\left\{
\begin{align*}
4(x - y) &= 500 \\
3(x + y) &= 465
\end{align*}
\right.
\]

The correct response is:
**{4(x−y)=500, 3(x+y)=465}**.