To find the width of the border, let's denote the width of the border as \( x \) inches. The dimensions of the framed photo will then 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:
\[ (10 + 2x)(6 + 2x) = 96 \]
Expanding this equation:
\[ 60 + 20x + 12x + 4x^2 = 96 \]
Combining like terms gives:
\[ 4x^2 + 32x + 60 = 96 \]
Subtracting 96 from both sides:
\[ 4x^2 + 32x - 36 = 0 \]
Dividing the entire equation by 4 to simplify:
\[ x^2 + 8x - 9 = 0 \]
Now we can factor the quadratic equation:
\[ (x + 9)(x - 1) = 0 \]
Setting each factor equal to zero gives us:
\[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
Since the width of the border cannot be negative, we have:
\[ x = 1 \]
So the width of the border is 1 inch.