To solve this problem, let's define some variables:
- Let \( b \) be the number of board books Santiago buys.
- Let \( s \) be the number of soft cover books he buys.
From the problem, we know that:
- The cost of board books is $1.50 each.
- The cost of soft cover books is $0.50 each.
- Santiago spends a total of $20.
We can set up the following equation based on the costs: \[ 1.5b + 0.5s = 20 \]
To simplify the equation, we can multiply everything by 2 to eliminate the decimals: \[ 3b + s = 40 \]
Now, we want to find solutions for \( b \) and \( s \) where both values are non-negative integers (since they can't purchase a negative number of books). Additionally, we need to make sure both types of books are included, which means \( b \geq 1 \) and \( s \geq 1 \).
Next, we can express \( s \) in terms of \( b \): \[ s = 40 - 3b \]
To ensure that \( s \) is a non-negative integer, we set up the following condition: \[ 40 - 3b \geq 1 \]
This gives: \[ 39 \geq 3b \] \[ 13 \geq b \]
So, \( b \) can take any integer value from 1 to 13 (inclusive).
Now, let's calculate the corresponding values of \( s \) for each value of \( b \):
- If \( b = 1 \): \( s = 40 - 3(1) = 37 \)
- If \( b = 2 \): \( s = 40 - 3(2) = 34 \)
- If \( b = 3 \): \( s = 40 - 3(3) = 31 \)
- If \( b = 4 \): \( s = 40 - 3(4) = 28 \)
- If \( b = 5 \): \( s = 40 - 3(5) = 25 \)
- If \( b = 6 \): \( s = 40 - 3(6) = 22 \)
- If \( b = 7 \): \( s = 40 - 3(7) = 19 \)
- If \( b = 8 \): \( s = 40 - 3(8) = 16 \)
- If \( b = 9 \): \( s = 40 - 3(9) = 13 \)
- If \( b = 10 \): \( s = 40 - 3(10) = 10 \)
- If \( b = 11 \): \( s = 40 - 3(11) = 7 \)
- If \( b = 12 \): \( s = 40 - 3(12) = 4 \)
- If \( b = 13 \): \( s = 40 - 3(13) = 1 \)
All of these combinations of \( (b, s) \) yield valid non-negative integers with both \( b \geq 1 \) and \( s \geq 1 \).
Thus, the number of viable solutions that include both board books and soft cover books is 13.
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