Questions: Solve the given system of equations by determinants. All numbers are approximate. 1.8y + 10.3 = -8.6x 3.4x + 4.6y + 12.5 = 0 x = , y = (Round to two decimal places as needed.)

Solution Steps

To solve the given system of equations using determinants, we can apply Cramer's Rule. This involves calculating the determinant of the coefficient matrix and using it to find the determinants for the numerators of \(x\) and \(y\). The solutions for \(x\) and \(y\) are then obtained by dividing these determinants by the determinant of the coefficient matrix.

We start with the given system of equations: \[ \begin{aligned} 1.8y + 10.3 &= -8.6x \\ 3.4x + 4.6y + 12.5 &= 0 \end{aligned} \] Rearranging these equations, we can express them in standard form: \[ \begin{aligned} 8.6x + 1.8y &= -10.3 \quad (1) \\ 3.4x + 4.6y &= -12.5 \quad (2) \end{aligned} \]

The coefficient matrix \(A\) and the constant matrix \(B\) can be defined as follows: \[ A = \begin{bmatrix} -8.6 & 1.8 \\ 3.4 & 4.6 \end{bmatrix}, \quad B = \begin{bmatrix} -10.3 \\ -12.5 \end{bmatrix} \]

We calculate the determinant of the coefficient matrix \(A\): \[ \text{det}(A) = -45.68 \] Next, we calculate the determinants for \(x\) and \(y\): \[ A_x = \begin{bmatrix} -10.3 & 1.8 \\ -12.5 & 4.6 \end{bmatrix}, \quad \text{det}(A_x) = -24.88 \] \[ A_y = \begin{bmatrix} -8.6 & -10.3 \\ 3.4 & -12.5 \end{bmatrix}, \quad \text{det}(A_y) = 142.52 \]

Using Cramer's Rule, we find the values of \(x\) and \(y\): \[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-24.88}{-45.68} \approx 0.54 \] \[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{142.52}{-45.68} \approx -3.12 \]

Final Answer