Use Cramer’s rule to solve the system [latex]\left\{\begin{array}{l}2 x-3 y=-4 \\5 x+7 y=1\end{array}\right.[/latex]

The determinant of the coefficient matrix is [latex]|D|=\left|\begin{array}{rr}2 & -3 \\5 & 7\end{array}\right|=29[/latex] [latex]\text {Hence,} \quad x=\frac{\left|D_x\right|}{|D|}=\frac{-25}{29}, \quad y=\frac{\left|D_y\right|}{|D|}=\frac{22}{29} .[/latex] Thus, the system has the unique solution [latex]\left(-\frac{25}{29}, \frac{22}{29}\right)[/latex].

Question 8.9.4: Use Cramer’s rule to solve the system {2x-3y=-4 5x+7y=1...

Precalculus Functions and Graphs [1893897]

Use Cramer’s rule to solve the system

$\left\{\begin{array}{l}2 x-3 y=-4 \\5 x+7 y=1\end{array}\right.$

Question Data is a breakdown of the data given in the question above.

System of equations:

Equation 1: 2x – 3y = -4

Equation 2: 5x + 7y = 1

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Step 1:

Calculate the determinant of the coefficient matrix, denoted as |D|. In this case, the coefficient matrix is
[2 -3] [5 7]
Calculating the determinant, we get |D| = 2*7 - (-3)*5 = 29.

Step 2:

Calculate the determinant of the matrix obtained by replacing the first column of the coefficient matrix with the constants on the right side of the equations. This matrix is denoted as |Dx|. In this case, the matrix is
[-25 -3] [ 22 7]
Calculating the determinant, we get |Dx| = (-25)*7 - (-3)*22 = -25.

Step 3:

Calculate the determinant of the matrix obtained by replacing the second column of the coefficient matrix with the constants on the right side of the equations. This matrix is denoted as |Dy|. In this case, the matrix is
[2 -25] [5 22]
Calculating the determinant, we get |Dy| = 2*22 - (-25)*5 = 22.

Step 4:

Finally, we can find the solution by dividing the determinants |Dx| and |Dy| by the determinant |D|.
x = |Dx| / |D| = -25 / 29 = -25/29 y = |Dy| / |D| = 22 / 29 = 22/29