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E. Use your graph to help you list three viable solutions to the problem when 2.00x + 1. 00y = 30.

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To find viable solutions for the equation \(2.00x + 1.00y = 30\), we can solve for \(y\) in terms of \(x\):

\[ 1.00y = 30 - 2.00x \] \[ y = 30 - 2x \]

To identify three viable solutions, we can choose different values for \(x\) and compute the corresponding \(y\) values.

  1. Choose \(x = 0\): \[ y = 30 - 2(0) = 30 \] Solution: \((0, 30)\)

  2. Choose \(x = 10\): \[ y = 30 - 2(10) = 30 - 20 = 10 \] Solution: \((10, 10)\)

  3. Choose \(x = 15\): \[ y = 30 - 2(15) = 30 - 30 = 0 \] Solution: \((15, 0)\)

Thus, three viable solutions to the equation \(2.00x + 1.00y = 30\) are:

  1. \((0, 30)\)
  2. \((10, 10)\)
  3. \((15, 0)\)
F. Suppose Santiago bought 26 books. Can you figure out how many of each type of book he bought? Explain. Write a system of equations and solve algebraically. Use Demos Calculator to check your work by adding the equation x + y = 26 to the graph above to help you answer the question.

Let's define two variables to represent the number of each type of book Santiago bought:

  • Let \( x \) be the number of the first type of book.
  • Let \( y \) be the number of the second type of book.

From the information given, we can create the following system of equations:

  1. The total number of books equation: \[ x + y = 26 \]

  2. The equation that comes from the problem statement you provided earlier \(2.00x + 1.00y = 30\), which can also represent the prices or some other condition regarding the books.

So the system of equations is: \[ \begin{align*}

  1. & \quad x + y = 26 \quad \text{(1)} \
  2. & \quad 2x + y = 30 \quad \text{(2)} \end{align*} \]

Now, we will solve this system of equations algebraically.

Step 1: Solve Equation (1) for \(y\)

From equation (1): \[ y = 26 - x \quad \text{(3)} \]

Step 2: Substitute Equation (3) into Equation (2)

Substituting (3) into equation (2): \[ 2x + (26 - x) = 30 \]

Step 3: Simplify and solve for \(x\)

\[ 2x + 26 - x = 30 \] \[ x + 26 = 30 \] \[ x = 30 - 26 \] \[ x = 4 \]

Step 4: Substitute \(x\) back into Equation (3) to find \(y\)

Now, substitute \(x = 4\) back into equation (3): \[ y = 26 - 4 = 22 \]

Conclusion

Santiago bought:

  • \(x = 4\) books of the first type,
  • \(y = 22\) books of the second type.

Check the Results

Now, let's verify this solution with the second equation: \[ 2(4) + 22 = 8 + 22 = 30 \] The second equation is satisfied.

Use the Calculator to Graph

You can graph the two equations:

  1. \(x + y = 26\)
  2. \(2x + y = 30\)

The intersection point of these two lines represents the solution to the system, which in this case should be \((4, 22)\).

Final Answer

Santiago bought 4 books of the first type and 22 books of the second type.

Make is shorter please

Sure! Here's a concise solution:

Definition of Variables

  • Let \( x \) = number of the first type of book
  • Let \( y \) = number of the second type of book

System of Equations

  1. \( x + y = 26 \) (total books)
  2. \( 2x + y = 30 \) (price condition)

Solve the System

  1. From equation 1: \[ y = 26 - x \]

  2. Substitute into equation 2: \[ 2x + (26 - x) = 30 \] \[ 2x + 26 - x = 30 \] \[ x + 26 = 30 \quad \Rightarrow \quad x = 4 \]

  3. Find \( y \): \[ y = 26 - 4 = 22 \]

Conclusion

Santiago bought:

  • 4 books of the first type
  • 22 books of the second type

This solution satisfies both equations!