To solve for the price of one pound of apples, let’s define the price of one pound of apples as \( a \) and the price of one pound of cheese as \( c \).
Week 1:
- Apples: \( 2\frac{1}{4} \) pounds can be converted to an improper fraction: \[ 2\frac{1}{4} = \frac{9}{4} \text{ pounds} \]
- Cheese: \( $5.73 \)
Therefore, the total amount spent in week 1 is: \[ \frac{9}{4}a + 5.73 \]
Week 2:
- Apples: \( 1\frac{4}{5} \) pounds can also be converted to an improper fraction: \[ 1\frac{4}{5} = \frac{9}{5} \text{ pounds} \]
- Cheese: \( $6.54 \)
Thus, the total amount spent in week 2 is: \[ \frac{9}{5}a + 6.54 \]
Setting the total spends equal:
Since Greg realized he spent the exact same amount each week, we can set the two totals equal to each other: \[ \frac{9}{4}a + 5.73 = \frac{9}{5}a + 6.54 \]
Solving for \( a \):
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Rearrange the equation: \[ \frac{9}{4}a - \frac{9}{5}a = 6.54 - 5.73 \]
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Calculate the right side: \[ 6.54 - 5.73 = 0.81 \]
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Now, simplify the left side: To combine the fractions, find a common denominator for \( \frac{9}{4} \) and \( \frac{9}{5} \), which is \( 20 \): \[ \frac{9}{4} = \frac{45}{20} \quad \text{and} \quad \frac{9}{5} = \frac{36}{20} \]
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Now the equation looks like this: \[ \left(\frac{45}{20} - \frac{36}{20}\right)a = 0.81 \]
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Simplifying gives: \[ \frac{9}{20}a = 0.81 \]
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To find \( a \), multiply both sides by the reciprocal of \( \frac{9}{20} \): \[ a = 0.81 \times \frac{20}{9} \]
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Performing the multiplication: \[ a = \frac{16.2}{9} \approx 1.8 \]
Conclusion:
Therefore, the price of one pound of apples is approximately \( \boxed{1.80} \) dollars.