Let the length of the note card be \( l \) inches and the width be \( w \) inches. We know the following:
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The perimeter \( P \) of a rectangle is given by the formula: \[ P = 2l + 2w \] Given that the perimeter is 18 inches, we can write: \[ 2l + 2w = 18 \] Simplifying this, we can divide everything by 2: \[ l + w = 9 \quad \text{(1)} \]
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The area \( A \) of a rectangle is given by the formula: \[ A = l \times w \] Given that the area is 18 square inches, we can write: \[ lw = 18 \quad \text{(2)} \]
Now, we can use equation (1) to express \( w \) in terms of \( l \): \[ w = 9 - l \]
Next, we substitute \( w \) into equation (2): \[ l(9 - l) = 18 \] Expanding this, we get: \[ 9l - l^2 = 18 \] Rearranging it gives us: \[ l^2 - 9l + 18 = 0 \]
Now, we can use the quadratic formula to solve for \( l \): \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -9 \), and \( c = 18 \): \[ l = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \] Calculating the discriminant: \[ l = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2} \]
This gives us two potential solutions for \( l \):
- \( l = \frac{12}{2} = 6 \)
- \( l = \frac{6}{2} = 3 \)
So, the dimensions (length and width) could be:
- If \( l = 6 \): then \( w = 9 - 6 = 3 \)
- If \( l = 3 \): then \( w = 9 - 3 = 6 \)
Thus, the dimensions of the note card can be written as: \[ \boxed{6 \text{ inches by } 3 \text{ inches}} \]