Questions: Solve the following system of equations using determinants. 5x + 6y = 10 2x + 3y = 4 A. x=2, y=0 B. x=-2, y=0 C. x=5, y=-6 D. x=-5, y=-6 E. None of the above

Solution Steps

We start with the system of equations: \[ \begin{align_} 5x + 6y &= 10 \quad (1) \\ 2x + 3y &= 4 \quad (2) \end{align_} \]

The coefficient matrix \( A \) is given by: \[ A = \begin{bmatrix} 5 & 6 \\ 2 & 3 \end{bmatrix} \]

To find the determinant of matrix \( A \): \[ \left| A \right| = \left| \begin{matrix} 5 & 6 \\ 2 & 3 \end{matrix} \right| = 5 \cdot 3 - 6 \cdot 2 = 15 - 12 = 3 \] Thus, \[ \text{det}(A) = 3.00 \]

Next, we construct the matrices \( A_x \) and \( A_y \): \[ A_x = \begin{bmatrix} 10 & 6 \\ 4 & 3 \end{bmatrix}, \quad A_y = \begin{bmatrix} 5 & 10 \\ 2 & 4 \end{bmatrix} \]

To find the determinant of matrix \( A_x \): \[ \left| A_x \right| = \left| \begin{matrix} 10 & 6 \\ 4 & 3 \end{matrix} \right| = 10 \cdot 3 - 6 \cdot 4 = 30 - 24 = 6 \] Thus, \[ \text{det}(A_x) = 6.00 \]

To find the determinant of matrix \( A_y \): \[ \left| A_y \right| = \left| \begin{matrix} 5 & 10 \\ 2 & 4 \end{matrix} \right| = 5 \cdot 4 - 10 \cdot 2 = 20 - 20 = 0 \] Thus, \[ \text{det}(A_y) = 0.00 \]

Using Cramer's Rule, we calculate \( x \) and \( y \): \[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{6}{3} = 2 \] \[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{0}{3} = 0 \]

The solution to the system of equations is: \[ x = 2, \quad y = 0 \]

Final Answer