Let's denote the number of board books Santiago buys as \( b \) and the number of soft cover books as \( s \).
The cost of board books is $1.50 each, so the total cost for the board books is \( 1.50b \). The cost of soft cover books is $0.50 each, so the total cost for the soft cover books is \( 0.50s \).
According to the problem, the total amount Santiago spends is $20. We can set up the following equation based on this information:
\[ 1.50b + 0.50s = 20 \]
To simplify this equation, we can multiply everything by 2 to eliminate the decimals:
\[ 3b + s = 40 \]
Now, we have a linear equation in two variables. To find possible solutions, we can express \( s \) in terms of \( b \):
\[ s = 40 - 3b \]
Since both \( b \) and \( s \) represent quantities of books, they must both be non-negative integers.
Now we can find possible integer values for \( b \):
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If \( b = 0 \): \[ s = 40 - 3(0) = 40 \quad (\text{Total cost } = 0 \times 1.50 + 40 \times 0.50 = 20) \]
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If \( b = 1 \): \[ s = 40 - 3(1) = 37 \quad (\text{Total cost } = 1 \times 1.50 + 37 \times 0.50 = 20) \]
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If \( b = 2 \): \[ s = 40 - 3(2) = 34 \quad (\text{Total cost } = 2 \times 1.50 + 34 \times 0.50 = 20) \]
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If \( b = 3 \): \[ s = 40 - 3(3) = 31 \quad (\text{Total cost } = 3 \times 1.50 + 31 \times 0.50 = 20) \]
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If \( b = 4 \): \[ s = 40 - 3(4) = 28 \quad (\text{Total cost } = 4 \times 1.50 + 28 \times 0.50 = 20) \]
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If \( b = 5 \): \[ s = 40 - 3(5) = 25 \quad (\text{Total cost } = 5 \times 1.50 + 25 \times 0.50 = 20) \]
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If \( b = 6 \): \[ s = 40 - 3(6) = 22 \quad (\text{Total cost } = 6 \times 1.50 + 22 \times 0.50 = 20) \]
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If \( b = 7 \): \[ s = 40 - 3(7) = 19 \quad (\text{Total cost } = 7 \times 1.50 + 19 \times 0.50 = 20) \]
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If \( b = 8 \): \[ s = 40 - 3(8) = 16 \quad (\text{Total cost } = 8 \times 1.50 + 16 \times 0.50 = 20) \]
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If \( b = 9 \): \[ s = 40 - 3(9) = 13 \quad (\text{Total cost } = 9 \times 1.50 + 13 \times 0.50 = 20) \]
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If \( b = 10 \): \[ s = 40 - 3(10) = 10 \quad (\text{Total cost } = 10 \times 1.50 + 10 \times 0.50 = 20) \]
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If \( b = 11 \): \[ s = 40 - 3(11) = 7 \quad (\text{Total cost } = 11 \times 1.50 + 7 \times 0.50 = 20) \]
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If \( b = 12 \): \[ s = 40 - 3(12) = 4 \quad (\text{Total cost } = 12 \times 1.50 + 4 \times 0.50 = 20) \]
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If \( b = 13 \): \[ s = 40 - 3(13) = 1 \quad (\text{Total cost } = 13 \times 1.50 + 1 \times 0.50 = 20) \]
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If \( b = 14 \): \[ s = 40 - 3(14) = -2 \quad (\text{Not allowed since }\ s \text{ cannot be negative}) \]
Now, we can list the valid integer pairs \( (b, s) \):
- \( (0, 40) \)
- \( (1, 37) \)
- \( (2, 34) \)
- \( (3, 31) \)
- \( (4, 28) \)
- \( (5, 25) \)
- \( (6, 22) \)
- \( (7, 19) \)
- \( (8, 16) \)
- \( (9, 13) \)
- \( (10, 10) \)
- \( (11, 7) \)
- \( (12, 4) \)
- \( (13, 1) \)
These pairs represent the different quantities of board books and soft cover books Santiago could buy with $20.
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