Answer :
Let's solve the given system of linear equations using Cramer's Rule. The system is:
[tex]\[ \begin{cases} 2x + 5y = -13 \\ -3x - 2y = 3 \end{cases} \][/tex]
To apply Cramer's Rule, we first need to find the determinant of the coefficient matrix [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 5 \\ -3 & -2 \end{pmatrix} \][/tex]
The determinant [tex]\(\det(A)\)[/tex] is given by:
[tex]\[ \det(A) = (2 \times -2) - (5 \times -3) \][/tex]
Calculating the terms within the determinant:
[tex]\[ \det(A) = -4 + 15 = 11 \][/tex]
Next, we need to find the determinant of matrix [tex]\(A_y\)[/tex], which is obtained by replacing the [tex]\(y\)[/tex]-coefficients in the original matrix with the constants from the right-hand side of the equations:
[tex]\[ A_y = \begin{pmatrix} 2 & -13 \\ -3 & 3 \end{pmatrix} \][/tex]
We calculate the determinant [tex]\(\det(A_y)\)[/tex]:
[tex]\[ \det(A_y) = (2 \times 3) - (-13 \times -3) \][/tex]
Calculating the terms within the determinant:
[tex]\[ \det(A_y) = 6 - 39 = -33 \][/tex]
Using Cramer's Rule, the value of [tex]\(y\)[/tex] is given by:
[tex]\[ y = \frac{\det(A_y)}{\det(A)} \][/tex]
Substituting the values of the determinants:
[tex]\[ y = \frac{-33}{11} = -3 \][/tex]
Therefore, the value of [tex]\(y\)[/tex] in the solution to the system of linear equations is [tex]\(-3\)[/tex]. The correct answer is:
[tex]\[ \boxed{-3} \][/tex]
[tex]\[ \begin{cases} 2x + 5y = -13 \\ -3x - 2y = 3 \end{cases} \][/tex]
To apply Cramer's Rule, we first need to find the determinant of the coefficient matrix [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 5 \\ -3 & -2 \end{pmatrix} \][/tex]
The determinant [tex]\(\det(A)\)[/tex] is given by:
[tex]\[ \det(A) = (2 \times -2) - (5 \times -3) \][/tex]
Calculating the terms within the determinant:
[tex]\[ \det(A) = -4 + 15 = 11 \][/tex]
Next, we need to find the determinant of matrix [tex]\(A_y\)[/tex], which is obtained by replacing the [tex]\(y\)[/tex]-coefficients in the original matrix with the constants from the right-hand side of the equations:
[tex]\[ A_y = \begin{pmatrix} 2 & -13 \\ -3 & 3 \end{pmatrix} \][/tex]
We calculate the determinant [tex]\(\det(A_y)\)[/tex]:
[tex]\[ \det(A_y) = (2 \times 3) - (-13 \times -3) \][/tex]
Calculating the terms within the determinant:
[tex]\[ \det(A_y) = 6 - 39 = -33 \][/tex]
Using Cramer's Rule, the value of [tex]\(y\)[/tex] is given by:
[tex]\[ y = \frac{\det(A_y)}{\det(A)} \][/tex]
Substituting the values of the determinants:
[tex]\[ y = \frac{-33}{11} = -3 \][/tex]
Therefore, the value of [tex]\(y\)[/tex] in the solution to the system of linear equations is [tex]\(-3\)[/tex]. The correct answer is:
[tex]\[ \boxed{-3} \][/tex]