Question

solve each system using cramer’s rule.

1. ( x - 4y = 1 )

( 2x + 3y = 13 )

1. ( 3x - 8y = -43 )

( -9x - 6y = -21 )

1. ( 6x - 2y = 36 )

( 5x + 4y = 47 )

1. ( 15x - 6y = -15 )

( 9x + 12y = 4 )

Explanation:

Problem 21: \( \boldsymbol{x - 4y = 1}\), \( \boldsymbol{2x + 3y = 13}\)

Step 1: Find the coefficient matrix determinant (\(D\))

The system is \(

$$\begin{cases} x - 4y = 1 \\ 2x + 3y = 13 \end{cases}$$

\). The coefficient matrix is \(

$$\begin{bmatrix} 1 & -4 \\ 2 & 3 \end{bmatrix}$$

\).
\( D = (1)(3) - (-4)(2) = 3 + 8 = 11 \).

Step 2: Find \(D_x\) (replace \(x\)-column with constants)

Constants: \(

$$\begin{bmatrix} 1 \\ 13 \end{bmatrix}$$

\). New matrix: \(

$$\begin{bmatrix} 1 & -4 \\ 13 & 3 \end{bmatrix}$$

\).
\( D_x = (1)(3) - (-4)(13) = 3 + 52 = 55 \).

Step 3: Find \(D_y\) (replace \(y\)-column with constants)

New matrix: \(

$$\begin{bmatrix} 1 & 1 \\ 2 & 13 \end{bmatrix}$$

\).
\( D_y = (1)(13) - (1)(2) = 13 - 2 = 11 \).

Step 4: Solve for \(x\) and \(y\)

\( x = \frac{D_x}{D} = \frac{55}{11} = 5 \), \( y = \frac{D_y}{D} = \frac{11}{11} = 1 \).

Step 1: Find \(D\)

Coefficient matrix: \(

$$\begin{bmatrix} 3 & -8 \\ -9 & -6 \end{bmatrix}$$

\).
\( D = (3)(-6) - (-8)(-9) = -18 - 72 = -90 \).

Step 2: Find \(D_x\) (replace \(x\)-column with \([-43, -21]\))

Matrix: \(

$$\begin{bmatrix} -43 & -8 \\ -21 & -6 \end{bmatrix}$$

\).
\( D_x = (-43)(-6) - (-8)(-21) = 258 - 168 = 90 \).

Step 3: Find \(D_y\) (replace \(y\)-column with \([-43, -21]\))

Matrix: \(

$$\begin{bmatrix} 3 & -43 \\ -9 & -21 \end{bmatrix}$$

\).
\( D_y = (3)(-21) - (-43)(-9) = -63 - 387 = -450 \).

Step 4: Solve for \(x\) and \(y\)

\( x = \frac{D_x}{D} = \frac{90}{-90} = -1 \), \( y = \frac{D_y}{D} = \frac{-450}{-90} = 5 \).

Step 1: Find \(D\)

Coefficient matrix: \(

$$\begin{bmatrix} 6 & -2 \\ 5 & 4 \end{bmatrix}$$

\).
\( D = (6)(4) - (-2)(5) = 24 + 10 = 34 \).

Step 2: Find \(D_x\) (replace \(x\)-column with \([36, 47]\))

Matrix: \(

$$\begin{bmatrix} 36 & -2 \\ 47 & 4 \end{bmatrix}$$

\).
\( D_x = (36)(4) - (-2)(47) = 144 + 94 = 238 \).

Step 3: Find \(D_y\) (replace \(y\)-column with \([36, 47]\))

Matrix: \(

$$\begin{bmatrix} 6 & 36 \\ 5 & 47 \end{bmatrix}$$

\).
\( D_y = (6)(47) - (36)(5) = 282 - 180 = 102 \).

Step 4: Solve for \(x\) and \(y\)

\( x = \frac{D_x}{D} = \frac{238}{34} = 7 \), \( y = \frac{D_y}{D} = \frac{102}{34} = 3 \).

Answer:

\( x = 5 \), \( y = 1 \)

Problem 22: \( \boldsymbol{3x - 8y = -43}\), \( \boldsymbol{-9x - 6y = -21}\)