To determine the number of viable solutions for Santiago's purchase of board books and soft cover books, we can set up a mathematical equation based on the problem.
Let's define:
- \( b \) = number of board books
- \( s \) = number of soft cover books
From the information given, we know that:
- Each board book costs $1.50
- Each soft cover book costs $0.50
- Santiago spends a total of $20
We can set up the following equation based on the prices and total spent:
\[ 1.5b + 0.5s = 20 \]
To simplify calculations, multiply the entire equation by 2 to eliminate decimal fractions:
\[ 3b + s = 40 \]
Now, we need to express the number of solutions involving both board books and soft cover books. First, we can express \( s \) in terms of \( b \):
\[ s = 40 - 3b \]
Now we need to find the constraints on \( b \) and \( s \). Since both \( b \) and \( s \) must be non-negative integers, we can evaluate the conditions:
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Since \( s \) must be non-negative: \[ 40 - 3b \geq 0 \implies 40 \geq 3b \implies \frac{40}{3} \geq b \implies b \leq 13.33 \] Since \( b \) must be an integer, the maximum value for \( b \) is 13.
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Since \( b \) must also be non-negative: \[ b \geq 0 \]
Now we can find the possible integer values for \( b \):
- The possible integer values for \( b \) are \( 0, 1, 2, \ldots, 13 \).
Next, we find the corresponding \( s \) values to ensure \( s \) is also a non-negative integer:
- If \( b = 0 \), \( s = 40 - 3(0) = 40 \)
- 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 \( b = 13 \), \( s = 40 - 3(13) = 1 \)
Now we need to ensure that each solution includes at least one of each type of book (both board and soft cover):
- Board books requirement: \( b \geq 1 \)
- Soft cover books requirement: \( s \geq 1 \)
The viable values for \( b \) satisfying both conditions (and \( s \geq 1 \)) are:
- \( b = 1 \): \( s = 37 \) (valid)
- \( b = 2 \): \( s = 34 \) (valid)
- \( b = 3 \): \( s = 31 \) (valid)
- \( b = 4 \): \( s = 28 \) (valid)
- \( b = 5 \): \( s = 25 \) (valid)
- \( b = 6 \): \( s = 22 \) (valid)
- \( b = 7 \): \( s = 19 \) (valid)
- \( b = 8 \): \( s = 16 \) (valid)
- \( b = 9 \): \( s = 13 \) (valid)
- \( b = 10 \): \( s = 10 \) (valid)
- \( b = 11 \): \( s = 7 \) (valid)
- \( b = 12 \): \( s = 4 \) (valid)
- \( b = 13 \): \( s = 1 \) (valid)
Thus, the values of \( b \) from 1 to 13 give us:
- There are \( 13 - 1 + 1 = 13 \) choices for \( b \), all resulting in valid \( s \) satisfying both book requirements.
Therefore, the total number of viable solutions that include both board books and soft cover books is 13.
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