To solve this problem, we need to set up an equation based on the given information:

• The original rectangular garden has dimensions 15 m by 18 m.
• A path of equal width is built around the garden, and the total area of the path is 150 m².
• We are asked to find the width of the path.

Step 1: Define the variable

Let the width of the path be $x$ meters.

Step 2: Set up the dimensions of the new, larger rectangle

The path surrounds the garden, so it adds $x$ meters to each side of the garden. The new dimensions of the rectangle, including the path, will be:

• Length: $18 + 2x$ (since the path adds $x$ meters on both sides of the length).
• Width: $15 + 2x$ (since the path adds $x$ meters on both sides of the width).

Step 3: Calculate the area of the new, larger rectangle

The area of the larger rectangle is: $\text{Area of larger rectangle} = (18 + 2x)(15 + 2x)$

Step 4: Subtract the area of the original garden

The area of the garden without the path is: $\text{Area of the garden} = 18 \times 15 = 270 \, \text{m}^2$ The area of the path is the difference between the area of the larger rectangle and the area of the original garden: $\text{Area of the path} = (18 + 2x)(15 + 2x) - 270$ We are told that the area of the path is 150 m²: $(18 + 2x)(15 + 2x) - 270 = 150$

Step 5: Solve the equation

Expand the left-hand side: $(18 + 2x)(15 + 2x) = 18 \times 15 + 18 \times 2x + 15 \times 2x + 2x \times 2x$ $= 270 + 36x + 30x + 4x^2$ $= 270 + 66x + 4x^2$ Substitute this into the equation: $270 + 66x + 4x^2 - 270 = 150$ Simplifying: $66x + 4x^2 = 150$ Rearrange the equation: $4x^2 + 66x - 150 = 0$

Step 6: Solve the quadratic equation

To solve $4x^2 + 66x - 150 = 0$, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $4x^2 + 66x - 150 = 0$, we have $a = 4$, $b = 66$, and $c = -150$.

Substitute these values into the quadratic formula: $x = \frac{-66 \pm \sqrt{66^2 - 4(4)(-150)}}{2(4)}$ $x = \frac{-66 \pm \sqrt{4356 + 2400}}{8}$ $x = \frac{-66 \pm \sqrt{6756}}{8}$ $x = \frac{-66 \pm 82.2}{8}$

Now solve for the two possible values for $x$: $x = \frac{-66 + 82.2}{8} \quad \text{or} \quad x = \frac{-66 - 82.2}{8}$ $x = \frac{16.2}{8} = 2.025 \quad \text{or} \quad x = \frac{-148.2}{8} = -18.525$

Since the width of the path cannot be negative, we discard $x = -18.525$ and keep $x = 2.025$.

Final Answer:

The width of the path is 2.03 meters (rounded to three significant figures).

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Would you like a detailed breakdown of any step? Here are some follow-up questions:

1. How does the quadratic formula work in this context?
2. What would happen if the area of the path was larger or smaller?
3. How could you model this problem with a different shape, like a circular garden?
4. How would the problem change if the path had varying widths along different sides?
5. Can you explain how to solve quadratic equations in general?

Tip: Always check for physical meaning in your solutions. Negative values may be mathematically correct but don't always fit real-world contexts!