\begin{tabular}{l} Ms. Malone is writing a quiz with 17 questions where each question is worth 3,4 , or 5 points based on difficulty. \\ She wants the number of 4 -point questions to be 2 more than the number of 5 -point questions, and she wants \\ the quiz to be worth a total of 62 points. \\ \hline 10. Write a system of equations in standard form to model this situation. \\ \begin{tabular}{l} 11. Create a matrix equation. \\ 12. Solve the matrix equation. \\ \hline 13. How many 3,4 and 5 point questions could there be? \end{tabular} \\ \hline\end{tabular}

To model Ms. Malone's quiz situation, let's define variables: let \( x \) be the number of 3-point questions, \( y \) be the number of 4-point questions, and \( z \) be the number of 5-point questions. We can formulate the following system of equations: 1. The total number of questions: \( x + y + z = 17 \) 2. The total points scored: \( 3x + 4y + 5z = 62 \) 3. The relationship between 4-point and 5-point questions: \( y = z + 2 \) --- To create a matrix equation, first rewrite the equations in a standard form: 1. \( x + y + z = 17 \) 2. \( 3x + 4y + 5z = 62 \) 3. \( 0x + y - z = 2 \) This can be represented in matrix form as \( A \vec{v} = \vec{b} \), where: \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 4 & 5 \\ 0 & 1 & -1 \end{bmatrix}, \quad \vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 17 \\ 62 \\ 2 \end{bmatrix} \] --- To solve the matrix equation, we can use the inverse of matrix \( A \) or row reduction. Instead, let's solve the original system step-by-step: From \( y = z + 2 \), substitute \( y \) in the first two equations: 1. \( x + (z + 2) + z = 17 \) ⟹ \( x + 2z + 2 = 17 \) ⟹ \( x + 2z = 15 \) 2. Substitute \( y \) in the second equation: \( 3x + 4(z + 2) + 5z = 62 \) ⟹ \( 3x + 4z + 8 + 5z = 62 \) ⟹ \( 3x + 9z = 54 \) ⟹ \( 3x = 54 - 9z \) ⟹ \( x = 18 - 3z \) Now we have two new equations: 1. \( x + 2z = 15 \) 2. \( x = 18 - 3z \) Substituting \( x \) gives: \( (18 - 3z) + 2z = 15 \) ⟹ \( 18 - z = 15 \) ⟹ \( z = 3 \). Substituting \( z = 3 \) into \( y = z + 2 \): \( y = 3 + 2 = 5 \). Now substitute \( y = 5 \) into \( x + y + z = 17 \): \( x + 5 + 3 = 17 \) ⟹ \( x = 9 \). --- So, in conclusion, there are 9 three-point questions, 5 four-point questions, and 3 five-point questions!