To find the width of the border, we'll first calculate the area of the photo before the border is added, and then find out how much area is taken by the border.
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Calculate the area of the original photo: \[ \text{Area of the photo} = \text{Length} \times \text{Width} = 10 \text{ inches} \times 6 \text{ inches} = 60 \text{ square inches} \]
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The total area of the framed picture (photo + border) is given as 96 square inches.
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Calculate the area of the border: \[ \text{Area of the border} = \text{Total area} - \text{Area of the photo} = 96 \text{ square inches} - 60 \text{ square inches} = 36 \text{ square inches} \]
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Let's represent the width of the border as \( x \). The dimensions of the entire framed picture will then be:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
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The area of the entire framed picture can also be expressed as: \[ \text{Area of the framed picture} = (10 + 2x)(6 + 2x) \]
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Set this equal to the known total area: \[ (10 + 2x)(6 + 2x) = 96 \]
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Expand the left side: \[ 60 + 20x + 12x + 4x^2 = 96 \] \[ 4x^2 + 32x + 60 = 96 \]
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Move 96 to the left: \[ 4x^2 + 32x + 60 - 96 = 0 \] \[ 4x^2 + 32x - 36 = 0 \]
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Divide the equation by 4: \[ x^2 + 8x - 9 = 0 \]
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Factor the quadratic: \[ (x + 9)(x - 1) = 0 \]
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So the solutions for \( x \) are: \[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \quad (\text{not a valid solution}) \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
Thus, the width of the border is 1 inch.
Therefore, the answer is 1 inch.
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