Let's define the variables:
- Let \( b \) be the number of board books.
- Let \( s \) be the number of soft cover books.
From the problem, we have two equations based on the total number of books and the total cost:
- \( b + s = 34 \) (total number of books)
- \( 1.5b + 0.5s = 20 \) (total cost of books)
Now, we can simplify the second equation. Multiply the entire equation by 2 to eliminate the decimal points:
\( 3b + s = 40 \).
Now we have the following system of equations:
- \( b + s = 34 \)
- \( 3b + s = 40 \)
Next, we can subtract the first equation from the second:
\[ (3b + s) - (b + s) = 40 - 34 \]
This simplifies to:
\[ 2b = 6 \]
Dividing both sides by 2 gives:
\[ b = 3 \]
Now substitute \( b = 3 \) back into the first equation to find \( s \):
\[ 3 + s = 34 \]
Solving for \( s \):
\[ s = 34 - 3 = 31 \]
Thus, Santiago purchased 3 board books and 31 soft cover books.
Now we can check the options provided:
- 17 board books and 17 soft cover books (not possible)
- 3 board books and 31 soft cover books (correct)
- 20 board books and 14 soft cover books (not possible)
- 15 board books and 19 soft cover books (not possible)
Therefore, the answer is:
3 board books and 31 soft cover books.