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To solve this problem, we need to determine the dimensions of the rectangular enclosure based on the given constraints: 

- The perimeter is 340 meters.
- The area is 6,000 m².

Let the length of the rectangle be \( L \) and the width be \( W \).

### Step 1: Perimeter Equation
The formula for the perimeter of a rectangle is:

\[
P = 2L + 2W
\]

We know the perimeter is 340 meters, so:

\[
340 = 2L + 2W
\]

Dividing by 2 to simplify:

\[
170 = L + W \quad \text{(Equation 1)}
\]

### Step 2: Area Equation
The formula for the area of a rectangle is:

\[
A = L \times W
\]

We know the area is 6,000 m², so:

\[
6000 = L \times W \quad \text{(Equation 2)}
\]

### Step 3: Solve the System of Equations
We now have two equations:

1. \( L + W = 170 \)
2. \( L \times W = 6000 \)

From Equation 1, solve for \( L \):

\[
L = 170 - W
\]

Substitute this into Equation 2:

\[
(170 - W) \times W = 6000
\]

Simplify:

\[
170W - W^2 = 6000
\]

Rearrange into a quadratic equation:

\[
W^2 - 170W + 6000 = 0
\]

### Step 4: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:

\[
W = \frac{-(-170) \pm \sqrt{(-170)^2 - 4(1)(6000)}}{2(1)}
\]

\[
W = \frac{170 \pm \sqrt{28900 - 24000}}{2}
\]

\[
W = \frac{170 \pm \sqrt{4900}}{2}
\]

\[
W = \frac{170 \pm 70}{2}
\]

So we have two possible solutions for \( W \):

\[
W = \frac{170 + 70}{2} = 120 \quad \text{or} \quad W = \frac{170 - 70}{2} = 50
\]

### Step 5: Find the Corresponding Length
Now, using \( L = 170 - W \):

- If \( W = 120 \), then \( L = 170 - 120 = 50 \).
- If \( W = 50 \), then \( L = 170 - 50 = 120 \).

### Final Dimensions
The dimensions of the enclosure can be either:

- Length = 120 meters and Width = 50 meters, or
- Length = 50 meters and Width = 120 meters.

These are the dimensions that Libby can use for her rectangular enclosure.

Let me know if you'd like more details or have any questions! Here are some additional questions for you:

1. What happens if the area or perimeter is changed?
2. How do you determine the optimal dimensions for other shapes?
3. What are the differences between perimeter and area?
4. How can you solve systems of equations involving real-world scenarios?
5. How does the quadratic formula help in solving problems with dimensions?

**Tip:** Always double-check your results, especially when interpreting dimensions and units in real-world problems!

Libby plans to create a rectangular pasturing enclosure. She has 340 m of fencing available for the enclosure’s perimeter and wants it to have an area of 6,000 m². What dimensions should Libby use?

Learn how to determine the dimensions of a rectangular enclosure with a perimeter of 340 meters and an area of 6,000 m² using algebra and quadratic equations. This step-by-step solution demonstrates solving a system of equations for length and width.

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