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.

Question

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

Answer 1

Let \( x \) be the number of board books Santiago buys, and \( y \) be the number of soft cover books Santiago buys.

The cost for board books is $1.50 each, so the total cost for board books is \( 1.50x \).

The cost for soft cover books is $0.50 each, so the total cost for soft cover books is \( 0.50y \).

Santiago spends a total of $20 on both types of books, which can be expressed with the following equation:

\[ 1.50x + 0.50y = 20 \]

This equation represents the relationship between the number of board books and soft cover books that Santiago buys and the total amount spent.

Answer 2

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.

Answer 3

To determine if Santiago could have bought 10 board books and 5 soft cover books, we can substitute \( x = 10 \) and \( y = 5 \) into the equation:

\[ 1.50x + 0.50y = 20 \]

Substituting in the values:

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

Calculating each term:

\[ 15 + 2.5 = 20 \]

Adding these amounts together:

\[ 17.5 \neq 20 \]

Since \( 17.5 \) does not equal \( 20 \), this means that it is not possible for Santiago to have bought 10 board books and 5 soft cover books within the $20 budget.

Thus, the answer is no, Santiago could not have bought that combination of books.

Answer 4

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.

Answer 5

To graph the equation \( 1.50x + 0.50y = 20 \), we'll rewrite it in slope-intercept form (\( y = mx + b \)).

1. Start with the original equation: \[ 1.50x + 0.50y = 20 \]

2. Rearrange to solve for \( y \): \[ 0.50y = 20 - 1.50x \] \[ y = \frac{20 - 1.50x}{0.50} \] \[ y = 40 - 3x \]

Now we can see that the slope (\( m \)) is -3 and the y-intercept (\( b \)) is 40.

Steps to Graph:

1. Plot the y-intercept: Start by plotting the point \( (0, 40) \) on the graph.

2. Use the slope: Since the slope is -3, this means that for every increase of 1 in \( x \), \( y \) decreases by 3. From the point \( (0, 40) \):

   • When \( x = 1 \): \[ y = 40 - 3(1) = 37 \quad \text{(Point (1, 37))} \]
   • When \( x = 2 \): \[ y = 40 - 3(2) = 34 \quad \text{(Point (2, 34))} \]
   • When \( x = 3 \): \[ y = 40 - 3(3) = 31 \quad \text{(Point (3, 31))} \]

3. Find another point: To ensure accuracy, you can use \( x = 10 \) to find another point:

   • When \( x = 10 \): \[ y = 40 - 3(10) = 10 \quad \text{(Point (10, 10))} \]

4. Plot points on the graph: Mark all the points you calculated:

   • \( (0, 40) \)
   • \( (1, 37) \)
   • \( (2, 34) \)
   • \( (3, 31) \)
   • \( (10, 10) \)

5. Draw the line: Connect the points with a straight line extending as much as needed.

Additional Notes:

• Make sure to extend the line in both directions to represent the equation fully.
• Keep in mind that in practical scenarios, \( x \) and \( y \) must be non-negative since they represent quantities of books.

Graphical Representation:

I can't create visual graphs for you directly, but following these steps will allow you to accurately graph the equation on graph paper. Remember to label the axes and indicate the line equation for clarity.

Answer 6

What are the constraints of the situation? Describe them.

Answer 7

In the context of the situation involving Santiago purchasing books for the day care, the main constraints will arise from the nature of the problem, particularly regarding the quantities of the books he can buy. Here are the key constraints to consider:

1. Non-negativity Constraints:

Both the number of board books and soft cover books must be non-negative integers. This means:

• \( x \geq 0 \) (number of board books)
• \( y \geq 0 \) (number of soft cover books)

Santiago cannot buy a negative number of books, and since he is purchasing whole books, \( x \) and \( y \) must be integers.

2. Budget Constraint:

Santiago has a maximum budget of $20 for purchasing books. This is already encapsulated in the equation \( 1.50x + 0.50y = 20 \). This means that the total cost of the books cannot exceed $20, which is reflected in any combination he chooses for \( x \) and \( y \).

3. Practical Constraints (if applicable):

While not mathematically defined like the above constraints, there may be practical considerations:

• Availability of Books: There may not be an unlimited supply of board books or soft cover books at the sale. If the sale has limited quantities of each, this could constrain the maximum values of \( x \) and \( y \).
• Desired Book Types: Santiago might have preferences for certain types of books that could further limit his options.

Summary of Constraints:

The constraints for Santiago are summarized as follows:

• \( x \geq 0 \) (number of board books)
• \( y \geq 0 \) (number of soft cover books)
• \( 1.50x + 0.50y \leq 20 \) (total cost of books must not exceed $20)

These constraints will limit the feasible combinations of board books and soft cover books that Santiago can purchase for the day care.

Answer 8

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

Answer 9

To find all viable solutions to the problem using the graph of the equation \( 1.50x + 0.50y = 20 \), we need to consider the constraints derived from the context:

1. Non-negativity constraints:

   • \( x \geq 0 \)
   • \( y \geq 0 \)

2. Budget constraint:

   • \( 1.50x + 0.50y \leq 20 \)

From the equation \( 1.50x + 0.50y = 20 \), we can begin to look for points where both \( x \) and \( y \) are integers and satisfy the equation while also abiding by the constraints.

Deriving Integer Solutions:

Rearranging:

We have already rewritten the original equation in slope-intercept form as: \[ y = 40 - 3x \]

Finding viable integer pairs \((x, y)\):

1. Start with integer values of \( x \), then calculate \( y \) while ensuring \( y \) is also a non-negative integer.

Evaluating Possible Values of \( x \):

• \( x = 0 \): \[ y = 40 - 3(0) = 40 \quad \Rightarrow \quad (0, 40) \]

• \( x = 1 \): \[ y = 40 - 3(1) = 37 \quad \Rightarrow \quad (1, 37) \]

• \( x = 2 \): \[ y = 40 - 3(2) = 34 \quad \Rightarrow \quad (2, 34) \]

• \( x = 3 \): \[ y = 40 - 3(3) = 31 \quad \Rightarrow \quad (3, 31) \]

• \( x = 4 \): \[ y = 40 - 3(4) = 28 \quad \Rightarrow \quad (4, 28) \]

• \( x = 5 \): \[ y = 40 - 3(5) = 25 \quad \Rightarrow \quad (5, 25) \]

• \( x = 6 \): \[ y = 40 - 3(6) = 22 \quad \Rightarrow \quad (6, 22) \]

• \( x = 7 \): \[ y = 40 - 3(7) = 19 \quad \Rightarrow \quad (7, 19) \]

• \( x = 8 \): \[ y = 40 - 3(8) = 16 \quad \Rightarrow \quad (8, 16) \]

• \( x = 9 \): \[ y = 40 - 3(9) = 13 \quad \Rightarrow \quad (9, 13) \]

• \( x = 10 \): \[ y = 40 - 3(10) = 10 \quad \Rightarrow \quad (10, 10) \]

• \( x = 11 \): \[ y = 40 - 3(11) = 7 \quad \Rightarrow \quad (11, 7) \]

• \( x = 12 \): \[ y = 40 - 3(12) = 4 \quad \Rightarrow \quad (12, 4) \]

• \( x = 13 \): \[ y = 40 - 3(13) = 1 \quad \Rightarrow \quad (13, 1) \]

• \( x = 14 \): \[ y = 40 - 3(14) = -2 \quad \Rightarrow \quad (14, -2) \quad \text{(not viable)} \]

List of Viable Integer Solutions:

Based on the calculations above, the viable solutions given the constraints are:

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

These points represent all possible combinations of board books (x) and soft cover books (y) that Santiago can purchase while staying within his $20 budget and not purchasing a negative number of books.

Answer 10

shorter

Answer 11

Here’s a shortened list of viable integer solutions for the problem, representing combinations of board books \( (x) \) and soft cover books \( (y) \), within the constraints \( x \geq 0 \) and \( y \geq 0 \):

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 pairs show how many board books and soft cover books Santiago can buy without exceeding his $20 budget.