Answer :
To solve the system of equations:
[tex]\[ \begin{array}{l} 5x + 2y = 9 \\ 2x - 3y = 15 \end{array} \][/tex]
we can use the method of Cramer's Rule, which is applicable because we have a system of linear equations with different slopes.
First, let's identify the coefficients and constants in the equations:
[tex]\[ \begin{array}{l} a_1 = 5, \, b_1 = 2, \, c_1 = 9 \\ a_2 = 2, \, b_2 = -3, \, c_2 = 15 \end{array} \][/tex]
According to Cramer's Rule, we first need to find the determinant [tex]\( D \)[/tex] of the coefficient matrix:
[tex]\[ D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{vmatrix} = (5 \cdot -3) - (2 \cdot 2) = -15 - 4 = -19 \][/tex]
Next, we find the determinant [tex]\( D_x \)[/tex] for the numerator of [tex]\( x \)[/tex] by replacing the [tex]\( x \)[/tex]-column with the constants:
[tex]\[ D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \\ \end{vmatrix} = (9 \cdot -3) - (15 \cdot 2) = -27 - 30 = -57 \][/tex]
Similarly, we find the determinant [tex]\( D_y \)[/tex] for the numerator of [tex]\( y \)[/tex] by replacing the [tex]\( y \)[/tex]-column with the constants:
[tex]\[ D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \\ \end{vmatrix} = (5 \cdot 15) - (2 \cdot 9) = 75 - 18 = 57 \][/tex]
The solutions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] can be found using:
[tex]\[ x = \frac{D_x}{D} = \frac{-57}{-19} = 3 \][/tex]
[tex]\[ y = \frac{D_y}{D} = \frac{57}{-19} = -3 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ \boxed{(3, -3)} \][/tex]
So, the correct answer is:
[tex]\[ \text{B. } (3, -3) \][/tex]
[tex]\[ \begin{array}{l} 5x + 2y = 9 \\ 2x - 3y = 15 \end{array} \][/tex]
we can use the method of Cramer's Rule, which is applicable because we have a system of linear equations with different slopes.
First, let's identify the coefficients and constants in the equations:
[tex]\[ \begin{array}{l} a_1 = 5, \, b_1 = 2, \, c_1 = 9 \\ a_2 = 2, \, b_2 = -3, \, c_2 = 15 \end{array} \][/tex]
According to Cramer's Rule, we first need to find the determinant [tex]\( D \)[/tex] of the coefficient matrix:
[tex]\[ D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{vmatrix} = (5 \cdot -3) - (2 \cdot 2) = -15 - 4 = -19 \][/tex]
Next, we find the determinant [tex]\( D_x \)[/tex] for the numerator of [tex]\( x \)[/tex] by replacing the [tex]\( x \)[/tex]-column with the constants:
[tex]\[ D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \\ \end{vmatrix} = (9 \cdot -3) - (15 \cdot 2) = -27 - 30 = -57 \][/tex]
Similarly, we find the determinant [tex]\( D_y \)[/tex] for the numerator of [tex]\( y \)[/tex] by replacing the [tex]\( y \)[/tex]-column with the constants:
[tex]\[ D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \\ \end{vmatrix} = (5 \cdot 15) - (2 \cdot 9) = 75 - 18 = 57 \][/tex]
The solutions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] can be found using:
[tex]\[ x = \frac{D_x}{D} = \frac{-57}{-19} = 3 \][/tex]
[tex]\[ y = \frac{D_y}{D} = \frac{57}{-19} = -3 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ \boxed{(3, -3)} \][/tex]
So, the correct answer is:
[tex]\[ \text{B. } (3, -3) \][/tex]