The price of one small packet and one big packet of sugar cane juice are RMx and RMy respectively. Haziq buys 6 small packets and 5 big packets of sugar cane juice at a total of RM19. The difference in price between one small packet and one big packet of the sugar cane juice is 50 sen. Using the matrix method, calculate the values of \( x \) and \( y \).

To solve the equations using the matrix method, first, we identify the equations based on the problem statement. 1. The total cost equation: \( 6x + 5y = 19 \) 2. The price difference equation: \( x - y = 0.5 \) Now, we can express this system of equations in matrix form \( Ax = B \). The coefficients matrix \( A \) and variable vector \( x \) can be set up as follows: \[ A = \begin{pmatrix} 6 & 5 \\ 1 & -1 \end{pmatrix}, \quad x = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 19 \\ 0.5 \end{pmatrix} \] Now, we can solve for \( x \) and \( y \) using the inverse of matrix \( A \). 1. First, calculate the determinant of \( A \): \[ \text{det}(A) = (6)(-1) - (5)(1) = -6 - 5 = -11 \] 2. Next, find the inverse of \( A \): \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} -1 & -5 \\ -1 & 6 \end{pmatrix} = \frac{1}{-11} \begin{pmatrix} -1 & -5 \\ -1 & 6 \end{pmatrix} = \begin{pmatrix} \frac{1}{11} & \frac{5}{11} \\ \frac{1}{11} & -\frac{6}{11} \end{pmatrix} \] 3. Now compute \( x = A^{-1}B \): \[ x = \begin{pmatrix} \frac{1}{11} & \frac{5}{11} \\ \frac{1}{11} & -\frac{6}{11} \end{pmatrix} \begin{pmatrix} 19 \\ 0.5 \end{pmatrix} \] Calculating the multiplication: \[ x_1 = \frac{1}{11}(19) + \frac{5}{11}(0.5) = \frac{19}{11} + \frac{2.5}{11} = \frac{21.5}{11} = 1.9545 \] \[ x_2 = \frac{1}{11}(19) - \frac{6}{11}(0.5) = \frac{19}{11} - \frac{3}{11} = \frac{16}{11} = 1.4545 \] Thus, approximately, \[ x \approx 1.95, \quad y \approx 1.45. \] In RM terms, the approximate prices are: **\(x \approx 1.95\) (for small packet)** and **\(y \approx 1.45\) (for big packet)**.