Consider the set of two linear equations: \[ x_{1}+2 x_{2}=5 \] \[ 1.5 x_{1}+9 x_{2}=1 \] Consider these equations in the matrix notation \( A X=B \). Use Cramer's rule to solve this set of equations. \( x_{1}=\underline{\text { Blank } 1}-2 \) decimal places \( x_{2}=\underline{\text { Blank } 2}-2 \) decimal places

To solve the system of equations using Cramer's Rule, we first represent the equations in matrix form \( AX = B \). The coefficient matrix \( A \) is given by: \[ A = \begin{pmatrix} 1 & 2 \\ 1.5 & 9 \end{pmatrix}, \] the variable matrix \( X \) is: \[ X = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, \] and the constant matrix \( B \) is: \[ B = \begin{pmatrix} 5 \\ 1 \end{pmatrix}. \] Next, we calculate the determinant of \( A \): \[ \text{det}(A) = (1)(9) - (2)(1.5) = 9 - 3 = 6. \] We calculate the determinants for \( D_1 \) and \( D_2 \): \[ D_1 = \begin{vmatrix} 5 & 2 \\ 1 & 9 \end{vmatrix} = (5)(9) - (2)(1) = 45 - 2 = 43, \] \[ D_2 = \begin{vmatrix} 1 & 5 \\ 1.5 & 1 \end{vmatrix} = (1)(1) - (5)(1.5) = 1 - 7.5 = -6.5. \] Now we can find \( x_1 \) and \( x_2 \) using Cramer's Rule: \[ x_1 = \frac{D_1}{\text{det}(A)} = \frac{43}{6} \approx 7.1667 \quad (\text{rounded to } 2 \text{ decimal places: } 7.17), \] \[ x_2 = \frac{D_2}{\text{det}(A)} = \frac{-6.5}{6} \approx -1.0833 \quad (\text{rounded to } 2 \text{ decimal places: } -1.08). \] Thus, the answers are: \( x_{1} = 7.17 \) \( x_{2} = -1.08 \)