A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of `120 m^(2)`. Find the width of the path.

Explanation

To solve the problem step by step, we will find the width of the path around the rectangular field.

### Step 1: Understand the dimensions of the rectangular field
The rectangular field has a length of 16 m and a width of 10 m.

### Step 2: Define the width of the path
Let the width of the path be \( x \) meters. The path surrounds the rectangular field uniformly.

### Step 3: Calculate the dimensions of the outer rectangle (field + path)
The length of the outer rectangle (including the path) will be:
\[ \text{Length} = 16 + 2x \]
The width of the outer rectangle (including the path) will be:
\[ \text{Width} = 10 + 2x \]

### Step 4: Calculate the area of the outer rectangle
The area of the outer rectangle is given by:
\[ \text{Area}_{\text{outer}} = (\text{Length}) \times (\text{Width}) = (16 + 2x)(10 + 2x) \]

### Step 5: Calculate the area of the rectangular field
The area of the rectangular field is:
\[ \text{Area}_{\text{field}} = 16 \times 10 = 160 \, \text{m}^2 \]

### Step 6: Set up the equation for the area of the path
The area of the path is given as 120 m². Therefore, we can express the area of the path as:
\[ \text{Area}_{\text{path}} = \text{Area}_{\text{outer}} - \text{Area}_{\text{field}} = 120 \]
Substituting the areas we calculated:
\[ (16 + 2x)(10 + 2x) - 160 = 120 \]

### Step 7: Simplify the equation
Expanding the left side:
\[ (16 + 2x)(10 + 2x) = 160 + 32x + 20x + 4x^2 = 160 + 52x + 4x^2 \]
So the equation becomes:
\[ 160 + 52x + 4x^2 - 160 = 120 \]
This simplifies to:
\[ 4x^2 + 52x - 120 = 0 \]

### Step 8: Divide the entire equation by 4
To simplify, divide the entire equation by 4:
\[ x^2 + 13x - 30 = 0 \]

### Step 9: Factor the quadratic equation
We need two numbers that multiply to -30 and add to 13. These numbers are 15 and -2.
So we can factor the equation as:
\[ (x + 15)(x - 2) = 0 \]

### Step 10: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x + 15 = 0 \) → \( x = -15 \) (not a valid solution since width cannot be negative)
2. \( x - 2 = 0 \) → \( x = 2 \)

### Conclusion
The width of the path is \( \boxed{2} \) meters.

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