To determine the width of the border, we start by calculating the area of the picture without the border.
The area of the picture is given by: \[ \text{Area} = \text{length} \times \text{width} = 10 , \text{inches} \times 6 , \text{inches} = 60 , \text{square inches}. \]
The total area of the entire framed picture, including the border, is given as 96 square inches.
Let \( x \) be the width of the border. The dimensions of the framed picture with the border will be:
- Length: \( 10 + 2x \) (adding the border on both sides)
- Width: \( 6 + 2x \) (adding the border on both sides)
The area of the framed picture can be expressed as: \[ \text{Total Area} = (10 + 2x)(6 + 2x). \] We set this equal to 96 square inches: \[ (10 + 2x)(6 + 2x) = 96. \]
Now, we expand the left-hand side: \[ 60 + 20x + 12x + 4x^2 = 96, \] which simplifies to: \[ 4x^2 + 32x + 60 = 96. \]
Subtract 96 from both sides: \[ 4x^2 + 32x + 60 - 96 = 0, \] which simplifies to: \[ 4x^2 + 32x - 36 = 0. \]
Dividing the entire equation by 4: \[ x^2 + 8x - 9 = 0. \]
Now we can factor the quadratic equation: \[ (x + 9)(x - 1) = 0. \]
This gives us two solutions: \[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \quad (\text{not valid}), \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1. \]
The valid width of the border is \( x = 1 \) inch.
Thus, the width of the border is 1 inch.