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To solve the system of linear equations:
[tex]\[ \begin{cases} -2x + 14y = 148 \\ 3x + 5y = 246 \end{cases} \][/tex]
we can use Cramer's Rule, which provides a straightforward way to solve for each variable. The general form of Cramer's Rule for a 2x2 system is:
[tex]\[ x = \frac{\Delta_x}{\Delta} \quad \text{and} \quad y = \frac{\Delta_y}{\Delta} \][/tex]
where [tex]\(\Delta\)[/tex] is the determinant of the coefficient matrix, [tex]\(\Delta_x\)[/tex] is the determinant of the matrix formed by replacing the [tex]\(x\)[/tex]-column with the constants from the right-hand side, and [tex]\(\Delta_y\)[/tex] is the determinant of the matrix formed by replacing the [tex]\(y\)[/tex]-column with the constants from the right-hand side.
Let's define the matrices involved:
1. The coefficient matrix [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} -2 & 14 \\ 3 & 5 \end{pmatrix} \][/tex]
2. The determinant [tex]\(\Delta\)[/tex] of matrix [tex]\(A\)[/tex]:
[tex]\[ \Delta = \left| \begin{array}{cc} -2 & 14 \\ 3 & 5 \end{array} \right| = (-2 \cdot 5) - (14 \cdot 3) = -10 - 42 = -52 \][/tex]
3. The matrix [tex]\(B_1\)[/tex] to solve for [tex]\(x\)[/tex], replacing the first column of [tex]\(A\)[/tex] with the constants:
[tex]\[ B_1 = \begin{pmatrix} 148 & 14 \\ 246 & 5 \end{pmatrix} \][/tex]
The determinant [tex]\(\Delta_x\)[/tex]:
[tex]\[ \Delta_x = \left| \begin{array}{cc} 148 & 14 \\ 246 & 5 \end{array} \right| = (148 \cdot 5) - (14 \cdot 246) = 740 - 3444 = -2704 \][/tex]
4. The matrix [tex]\(B_2\)[/tex] to solve for [tex]\(y\)[/tex], replacing the second column of [tex]\(A\)[/tex] with the constants:
[tex]\[ B_2 = \begin{pmatrix} -2 & 148 \\ 3 & 246 \end{pmatrix} \][/tex]
The determinant [tex]\(\Delta_y\)[/tex]:
[tex]\[ \Delta_y = \left| \begin{array}{cc} -2 & 148 \\ 3 & 246 \end{array} \right| = (-2 \cdot 246) - (148 \cdot 3) = -492 - 444 = -936 \][/tex]
Now, apply Cramer's Rule to find [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x = \frac{\Delta_x}{\Delta} = \frac{-2704}{-52} = 52 \][/tex]
[tex]\[ y = \frac{\Delta_y}{\Delta} = \frac{-936}{-52} = 18 \][/tex]
Thus, we can determine the [tex]\(x\)[/tex]-value of the solution to the system of linear equations using the equation:
[tex]\[ x = \frac{\left| \begin{array}{cc} 148 & 14 \\ 246 & 5 \end{array} \right|}{\left| \begin{array}{cc} -2 & 14 \\ 3 & 5 \end{array} \right|} = 52 \][/tex]
Therefore, the equation that can be used to determine the [tex]\(x\)[/tex]-value of the solution is:
[tex]\[ \frac{\left| \begin{array}{cc} 148 & 14 \\ 246 & 5 \end{array} \right|}{\left| \begin{array}{cc} -2 & 14 \\ 3 & 5 \end{array} \right|} = 52 \][/tex]
[tex]\[ \begin{cases} -2x + 14y = 148 \\ 3x + 5y = 246 \end{cases} \][/tex]
we can use Cramer's Rule, which provides a straightforward way to solve for each variable. The general form of Cramer's Rule for a 2x2 system is:
[tex]\[ x = \frac{\Delta_x}{\Delta} \quad \text{and} \quad y = \frac{\Delta_y}{\Delta} \][/tex]
where [tex]\(\Delta\)[/tex] is the determinant of the coefficient matrix, [tex]\(\Delta_x\)[/tex] is the determinant of the matrix formed by replacing the [tex]\(x\)[/tex]-column with the constants from the right-hand side, and [tex]\(\Delta_y\)[/tex] is the determinant of the matrix formed by replacing the [tex]\(y\)[/tex]-column with the constants from the right-hand side.
Let's define the matrices involved:
1. The coefficient matrix [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} -2 & 14 \\ 3 & 5 \end{pmatrix} \][/tex]
2. The determinant [tex]\(\Delta\)[/tex] of matrix [tex]\(A\)[/tex]:
[tex]\[ \Delta = \left| \begin{array}{cc} -2 & 14 \\ 3 & 5 \end{array} \right| = (-2 \cdot 5) - (14 \cdot 3) = -10 - 42 = -52 \][/tex]
3. The matrix [tex]\(B_1\)[/tex] to solve for [tex]\(x\)[/tex], replacing the first column of [tex]\(A\)[/tex] with the constants:
[tex]\[ B_1 = \begin{pmatrix} 148 & 14 \\ 246 & 5 \end{pmatrix} \][/tex]
The determinant [tex]\(\Delta_x\)[/tex]:
[tex]\[ \Delta_x = \left| \begin{array}{cc} 148 & 14 \\ 246 & 5 \end{array} \right| = (148 \cdot 5) - (14 \cdot 246) = 740 - 3444 = -2704 \][/tex]
4. The matrix [tex]\(B_2\)[/tex] to solve for [tex]\(y\)[/tex], replacing the second column of [tex]\(A\)[/tex] with the constants:
[tex]\[ B_2 = \begin{pmatrix} -2 & 148 \\ 3 & 246 \end{pmatrix} \][/tex]
The determinant [tex]\(\Delta_y\)[/tex]:
[tex]\[ \Delta_y = \left| \begin{array}{cc} -2 & 148 \\ 3 & 246 \end{array} \right| = (-2 \cdot 246) - (148 \cdot 3) = -492 - 444 = -936 \][/tex]
Now, apply Cramer's Rule to find [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x = \frac{\Delta_x}{\Delta} = \frac{-2704}{-52} = 52 \][/tex]
[tex]\[ y = \frac{\Delta_y}{\Delta} = \frac{-936}{-52} = 18 \][/tex]
Thus, we can determine the [tex]\(x\)[/tex]-value of the solution to the system of linear equations using the equation:
[tex]\[ x = \frac{\left| \begin{array}{cc} 148 & 14 \\ 246 & 5 \end{array} \right|}{\left| \begin{array}{cc} -2 & 14 \\ 3 & 5 \end{array} \right|} = 52 \][/tex]
Therefore, the equation that can be used to determine the [tex]\(x\)[/tex]-value of the solution is:
[tex]\[ \frac{\left| \begin{array}{cc} 148 & 14 \\ 246 & 5 \end{array} \right|}{\left| \begin{array}{cc} -2 & 14 \\ 3 & 5 \end{array} \right|} = 52 \][/tex]
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