Let the dimensions of the rectangular note card be \( l \) (length) and \( w \) (width).
We know two things:
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The perimeter \( P \) of the rectangle is given by: \[ P = 2l + 2w = 18 \] Simplifying this, we get: \[ l + w = 9 \quad \text{(Equation 1)} \]
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The area \( A \) of the rectangle is given by: \[ A = lw = 18 \quad \text{(Equation 2)} \]
Now we can express \( w \) in terms of \( l \) using Equation 1: \[ w = 9 - l \]
Next, we can substitute this expression for \( w \) into Equation 2: \[ l(9 - l) = 18 \]
Expanding this gives: \[ 9l - l^2 = 18 \] Rearranging the equation leads to: \[ l^2 - 9l + 18 = 0 \]
We can use the quadratic formula to solve for \( l \): \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = -9, c = 18 \): \[ l = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 1 \times 18}}{2 \times 1} \] Calculating the discriminant: \[ 9^2 - 4 \cdot 1 \cdot 18 = 81 - 72 = 9 \] Now substituting back: \[ l = \frac{9 \pm 3}{2} \]
This gives us two possible values for \( l \):
- \( l = \frac{12}{2} = 6 \)
- \( l = \frac{6}{2} = 3 \)
Now we can find the corresponding \( w \) for both values of \( l \):
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If \( l = 6 \): \[ w = 9 - 6 = 3 \]
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If \( l = 3 \): \[ w = 9 - 3 = 6 \]
Thus, the dimensions of the note card are \( 6 \) inches by \( 3 \) inches.
Therefore, the dimensions of the note card are: \[ \text{6 inches by 3 inches} \]