The price of one kaya bun is RMx and the price of one butter bun is RMy. Suzi buys 6 kaya buns and 3 butter buns while Fauzi buys 4 kaya buns and 5 butter buns. Suzi pays RM8.40 and Fauzi pays RM1 less than Suzi. Using the matrix method, calculate the price of one kaya bun and the price of one butter bun.
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Bonus Knowledge
To solve this problem using matrices, we first need to set up the equations based on the purchases made by Suzi and Fauzi. From the information given: 1. Suzi's purchase: \[ 6x + 3y = 8.40 \] 2. Fauzi's purchase: \[ 4x + 5y = 8.40 - 1 = 7.40 \] Now we can write the system of equations: \[ \begin{align*} 6x + 3y &= 8.40 \quad (1) \\ 4x + 5y &= 7.40 \quad (2) \end{align*} \] Next, we can express this system in matrix form: \[ \begin{bmatrix} 6 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8.40 \\ 7.40 \end{bmatrix} \] Now we can solve for \(x\) (the price of one kaya bun) and \(y\) (the price of one butter bun) using the inverse of the coefficient matrix. First, calculate the determinant of the coefficient matrix: \[ \text{Det} = (6)(5) - (3)(4) = 30 - 12 = 18 \] Now, find the inverse of the coefficient matrix: \[ \begin{bmatrix} 6 & 3 \\ 4 & 5 \end{bmatrix}^{-1} = \frac{1}{18} \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} \] Now we multiply this inverse matrix by the constants matrix: \[ \begin{bmatrix} 5 & -3 \\ -4 & 6 \end{bmatrix} \begin{bmatrix} 8.40 \\ 7.40 \end{bmatrix} = \begin{bmatrix} (5)(8.40) + (-3)(7.40) \\ (-4)(8.40) + (6)(7.40) \end{bmatrix} \] Calculating these: 1. For \(x\): \[ 5(8.40) - 3(7.40) = 42 - 22.20 = 19.80 \] 2. For \(y\): \[ -4(8.40) + 6(7.40) = -33.60 + 44.40 = 10.80 \] Now, dividing by the determinant (18): \[ x = \frac{19.80}{18} = 1.10 \] \[ y = \frac{10.80}{18} = 0.60 \] Therefore, the price of one kaya bun is **RM1.10** and the price of one butter bun is **RM0.60**.