A. Let \( b \) be the number of board books and \( s \) be the number of soft cover books. The two-variable equation representing the situation is:
\[ 1.50b + 0.50s = 20 \]
B. No, it is not possible that Santiago bought 10 board books and five soft cover books because substituting these values gives \( 1.50(10) + 0.50(5) = 15 + 2.5 = 17.5 \), which is less than $20.
C. The constraints of the situation are that the number of board books (\( b \)) and the number of soft cover books (\( s \)) must be non-negative integers (i.e., \( b \geq 0 \) and \( s \geq 0 \)) and their combined total price must equal $20.
D. To find all viable solutions, we can express \( s \) in terms of \( b \) using the original equation:
\[ s = 40 - 3b \]
The solutions must yield non-negative integers for both \( b \) and \( s \). Therefore, possible pairs \((b, s)\) that satisfy the equation with \( b \) being from 0 to 13 (as 14 board books alone would exceed $20) would be:
- \( (0, 40) \)
- \( (1, 37) \)
- \( (2, 34) \)
- \( (3, 31) \)
- \( (4, 28) \)
- \( (5, 25) \)
- \( (6, 22) \)
- \( (7, 19) \)
- \( (8, 16) \)
- \( (9, 13) \)
- \( (10, 10) \)
- \( (11, 7) \)
- \( (12, 4) \)
- \( (13, 1) \)
E. Yes, if Santiago bought 34 books, then he could be described using the system of equations:
\[ b + s = 34 \] and \[ 1.50b + 0.50s = 20 \]. By solving these equations simultaneously, we find \( b = 8 \) (board books) and \( s = 26 \) (soft cover books).