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)

To write a system of linear equations representing the scenario, we start by defining the effective speeds of the plane when flying against and with the wind.

1. **Flying against the wind**:
- The speed of the plane against the wind is \( x - y \).
- The distance flown is 500 miles.
- The time taken is 4 hours.
- Using the formula: distance = speed × time, we can set up the equation:
\[
4(x - y) = 500
\]

2. **Flying with the wind**:
- The speed of the plane with the wind is \( x + y \).
- The distance flown is 465 miles.
- The time taken is 3 hours.
- Again, using the distance = speed × time formula, we get:
\[
3(x + y) = 465
\]

Now we can rewrite these equations:

1. From the first equation:
\[
4x - 4y = 500
\]

2. From the second equation:
\[
3x + 3y = 465
\]

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

Now, let's look for the corresponding option from your responses:

- **First option:** \( \{4x + y = 500, 3x - y = 465\} \)
- **Second option:** \( \{4x - y = 500, 3x + y = 465\} \)
- **Third option:** \( \{4(x - y) = 500, 3(x + y) = 465\} \)
- **Fourth option:** \( \{4(x + y) = 500, 3(x - y) = 465\} \)

The equations \( 4x - 4y = 500 \) and \( 3x + 3y = 465 \) simplifies to \( 4(x - y) = 500 \) and \( 3(x + y) = 465\), which is option **3**.

So the correct answer is:
\[
\{4(x - y) = 500, 3(x + y) = 465\}
\]