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Santiago is working at his mom’s day care this summer. His mom has asked him to check out the library book sale to look for books they might be able to use in the day care. When he gets to the book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a variety of books and spends $20.

A. Write a two-variable equation to represent this situation. Be sure to define the variables you use in your equation.

11 answers

Let \( b \) represent the number of board books that Santiago buys, and let \( s \) represent the number of soft cover books he buys. The cost of each board book is $1.50, and the cost of each soft cover book is $0.50.

We can express the total amount he spends with the following equation:

\[ 1.50b + 0.50s = 20 \]

In this equation:

  • \( b \) = number of board books
  • \( s \) = number of soft cover books

This equation models the situation as it accounts for the total spending based on the price of each type of book.

B. Is it possible that Santiago bought 10 board books and five soft cover books? Use your equation from part A to check. Explain your answer.

To check if Santiago could have bought 10 board books and 5 soft cover books, we can substitute \( b = 10 \) and \( s = 5 \) into the equation from part A:

\[ 1.50b + 0.50s = 20 \]

Substituting the values:

\[ 1.50(10) + 0.50(5) = 20 \]

Calculating each term:

\[ 15 + 2.5 = 20 \]

Now add these amounts:

\[ 15 + 2.5 = 17.5 \]

Since \( 17.5 \) does not equal \( 20 \), it is not possible for Santiago to have bought 10 board books and 5 soft cover books.

Therefore, the answer is no; the scenario does not satisfy the equation, which indicates that the total spent would be less than the $20 he had.

C. Graph the solution set for the equation you wrote in part A on a piece of graph paper. Note: At this point, do not yet consider any constraints given by the situation.

To graph the equation \( 1.50b + 0.50s = 20 \), it is helpful to first rearrange it into slope-intercept form (i.e., \( s = mx + b \)).

  1. Start with the equation: \[ 1.50b + 0.50s = 20 \]

  2. Isolate \( s \): \[ 0.50s = 20 - 1.50b \]

  3. Divide everything by 0.50: \[ s = -3b + 40 \]

Now, we have \( s = -3b + 40 \). This equation represents a line with a slope of -3 and a y-intercept of 40.

Steps to Graph:

  1. Find the y-intercept:

    • When \( b = 0 \): \[ s = -3(0) + 40 = 40 \]
    • So, the point (0, 40) is on the graph.

  2. Find the x-intercept:

    • When \( s = 0 \): \[ 0 = -3b + 40 \implies 3b = 40 \implies b = \frac{40}{3} \approx 13.33 \]
    • So, the point \((\frac{40}{3}, 0) \approx (13.33, 0)\) is on the graph.

  3. Plot the points:

    • Plot the points (0, 40) and approximately (13.33, 0) on the graph.

  4. Draw the line:

    • Draw a straight line through these points, extending in both directions.

Important Note:

Since the context of the problem involves the number of books, only non-negative integer values make sense for \( b \) and \( s \). Therefore, when interpreting the graph, we restrict the values of \( b \) and \( s \) to the first quadrant where both variables are non-negative.

Result:

You will see a straight line that begins at (0, 40) and goes down to approximately (13.33, 0). The feasible solution area (for the book scenario) would be the portion of this line that lies within the first quadrant (where both \( b \) and \( s \) are greater than or equal to 0).

If you need any further assistance with the graphing process or specific graph paper instructions, please let me know!

D. What are the constraints of the situation? Describe them.

In the context of the situation involving Santiago's book purchases for the day care, there are a few constraints that we need to consider. Here are the main constraints:

  1. Non-negativity Constraint:

    • Santiago cannot buy a negative number of books. Therefore, both the number of board books (\( b \)) and the number of soft cover books (\( s \)) must be non-negative integers: \[ b \geq 0 \quad \text{and} \quad s \geq 0 \]

  2. Budget Constraint:

    • Throughout the book sale, Santiago has a maximum budget of $20. This is already represented in the equation \( 1.50b + 0.50s = 20 \). This means that any combination of board books and soft cover books must not exceed this budget.

  3. Book Type Limitations:

    • As he is purchasing books specifically for a day care, it is reasonable to assume there is likely a limit to the practical number of books one can buy. Realistically, Santiago wouldn't be able to buy an infinite number of books due to space, usability, or other resource limitations.

  4. Integers Only:

    • Since Santiago cannot purchase a fraction of a book, both \( b \) and \( s \) must be whole numbers (integers): \[ b \in \mathbb{Z}^+ \quad \text{and} \quad s \in \mathbb{Z}^+ \]

These constraints define the feasible region for \( b \) and \( s \) in relation to the equation \( 1.50b + 0.50s = 20 \). Graphically, this means looking for integer coordinates in the first quadrant that lie on or below the line represented by the equation, within the given conditions.

In summary, the constraints ensure that the number of books is realistic, within the budget, and logically applicable to the context of a day care situation.

E. Use your graph to help you make a list of all viable solutions to the problem.

To find all viable solutions to the problem using the graph of the equation \( 1.50b + 0.50s = 20 \) along with the constraints we identified (non-negativity and integers only), we'll look for integer pairs \((b, s)\) that satisfy the equation while also being greater than or equal to zero.

  1. Revisiting the Equation:

    • The equation is \( 1.50b + 0.50s = 20 \).
    • We can rewrite it to isolate \( s \): \[ s = -3b + 40 \]

  2. Finding Integer Solutions:

    • We can calculate values for \( s \) by substituting different non-negative integer values for \( b \).

Calculating Solutions:

Let’s find the integer values by substituting successive values for \( b \):

  • For \( b = 0 \): \[ s = -3(0) + 40 = 40 \quad \Rightarrow \quad (0, 40) \]

  • For \( b = 1 \): \[ s = -3(1) + 40 = 37 \quad \Rightarrow \quad (1, 37) \]

  • For \( b = 2 \): \[ s = -3(2) + 40 = 34 \quad \Rightarrow \quad (2, 34) \]

  • For \( b = 3 \): \[ s = -3(3) + 40 = 31 \quad \Rightarrow \quad (3, 31) \]

  • For \( b = 4 \): \[ s = -3(4) + 40 = 28 \quad \Rightarrow \quad (4, 28) \]

  • For \( b = 5 \): \[ s = -3(5) + 40 = 25 \quad \Rightarrow \quad (5, 25) \]

  • For \( b = 6 \): \[ s = -3(6) + 40 = 22 \quad \Rightarrow \quad (6, 22) \]

  • For \( b = 7 \): \[ s = -3(7) + 40 = 19 \quad \Rightarrow \quad (7, 19) \]

  • For \( b = 8 \): \[ s = -3(8) + 40 = 16 \quad \Rightarrow \quad (8, 16) \]

  • For \( b = 9 \): \[ s = -3(9) + 40 = 13 \quad \Rightarrow \quad (9, 13) \]

  • For \( b = 10 \): \[ s = -3(10) + 40 = 10 \quad \Rightarrow \quad (10, 10) \]

  • For \( b = 11 \): \[ s = -3(11) + 40 = 7 \quad \Rightarrow \quad (11, 7) \]

  • For \( b = 12 \): \[ s = -3(12) + 40 = 4 \quad \Rightarrow \quad (12, 4) \]

  • For \( b = 13 \): \[ s = -3(13) + 40 = 1 \quad \Rightarrow \quad (13, 1) \]

  • For \( b = 14 \): \[ s = -3(14) + 40 = -2 \quad \Rightarrow \quad \text{(invalid since \( s < 0 \))} \]

Viable Solutions List:

The list of viable integer solutions \((b, s)\) that Santiago can choose is:

  1. \( (0, 40) \)
  2. \( (1, 37) \)
  3. \( (2, 34) \)
  4. \( (3, 31) \)
  5. \( (4, 28) \)
  6. \( (5, 25) \)
  7. \( (6, 22) \)
  8. \( (7, 19) \)
  9. \( (8, 16) \)
  10. \( (9, 13) \)
  11. \( (10, 10) \)
  12. \( (11, 7) \)
  13. \( (12, 4) \)
  14. \( (13, 1) \)

These solutions represent all the combinations of board books \( b \) and soft cover books \( s \) that Santiago can buy without exceeding his budget while also sticking to the constraints of the problem.

F. Suppose Santiago bought 34 books. Can you figure out how many of each type of book he bought? Explain.

To determine how many of each type of book Santiago bought if he purchased a total of 34 books, we need to set up a system of equations using the information we have.

  1. Define Variables:

    • Let \( b \) be the number of board books.
    • Let \( s \) be the number of soft cover books.

  2. Set up the equations: We have two relationships based on the problem:

    • The total number of books: \[ b + s = 34 \]
    • The total cost: \[ 1.50b + 0.50s = 20 \]

  3. Substituting for one variable: From the first equation, we can express \( s \) in terms of \( b \): \[ s = 34 - b \]

  4. Substituting into the cost equation: Substitute \( s \) in the cost equation: \[ 1.50b + 0.50(34 - b) = 20 \]

  5. Distributing and simplifying: \[ 1.50b + 17 - 0.50b = 20 \] \[ (1.50 - 0.50)b + 17 = 20 \] \[ 1.00b + 17 = 20 \]

  6. Solving for \( b \): \[ 1.00b = 20 - 17 \] \[ 1.00b = 3 \quad \Rightarrow \quad b = 3 \]

  7. Finding \( s \): Now substitute \( b = 3 \) back into the equation for \( s \): \[ s = 34 - b = 34 - 3 = 31 \]

Conclusion:

Therefore, if Santiago bought 34 books in total, he purchased:

  • Board Books: \( b = 3 \)
  • Soft Cover Books: \( s = 31 \)

You can verify the solution by checking that the total cost matches: \[ 1.50 \times 3 + 0.50 \times 31 = 4.50 + 15.50 = 20 \]

This confirms that the numbers are correct.