Solved on Jan 24, 2024

Solve the linear system using Cramer's Rule. Find the determinants $D$ , $D_x$ , and $D_y$ . The solution is $(x, y) = ($ $\frac{D_x}{D}$ $,$ $\frac{D_y}{D}$ $)$ .

STEP 1

Assumptions
1. We are given a system of linear equations: $\begin{array}{l} -3 x-y=-11 \\ 2 x+4 y=-6 \end{array}$
2. We will use Cramer's Rule to solve for $x$ and $y$ .
3. Cramer's Rule states that for a system of two linear equations: $\begin{array}{l} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{array}$ the solution can be found using the following formulas, provided that the determinant $D$ is not zero: $x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$ where $D$ is the determinant of the coefficient matrix, $D_x$ is the determinant of the matrix formed by replacing the $x$ -coefficients with the constants, and $D_y$ is the determinant of the matrix formed by replacing the $y$ -coefficients with the constants.

STEP 2

Calculate the determinant $D$ of the coefficient matrix. The coefficient matrix is: $\begin{pmatrix} -3 & -1 \\ 2 & 4 \end{pmatrix}$ The determinant $D$ is calculated as: $D = (-3)(4) - (-1)(2)$

STEP 3

Compute the value of the determinant $D$ . $D = (-3)(4) - (-1)(2) = -12 - (-2) = -12 + 2 = -10$

STEP 4

Calculate the determinant $D_x$ by replacing the $x$ -coefficients with the constants from the right-hand side of the equations. The modified matrix is: $\begin{pmatrix} -11 & -1 \\ -6 & 4 \end{pmatrix}$ The determinant $D_x$ is calculated as: $D_x = (-11)(4) - (-1)(-6)$

STEP 5

Compute the value of the determinant $D_x$ . $D_x = (-11)(4) - (-1)(-6) = -44 - 6 = -50$

STEP 6

Calculate the determinant $D_y$ by replacing the $y$ -coefficients with the constants from the right-hand side of the equations. The modified matrix is: $\begin{pmatrix} -3 & -11 \\ 2 & -6 \end{pmatrix}$ The determinant $D_y$ is calculated as: $D_y = (-3)(-6) - (2)(-11)$

STEP 7

Compute the value of the determinant $D_y$ . $D_y = (-3)(-6) - (2)(-11) = 18 + 22 = 40$

STEP 8

Now that we have the determinants $D$ , $D_x$ , and $D_y$ , we can find the values of $x$ and $y$ using Cramer's Rule. $x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$

STEP 9

Calculate the value of $x$ . $x = \frac{D_x}{D} = \frac{-50}{-10} = 5$

STEP 10

Calculate the value of $y$ . $y = \frac{D_y}{D} = \frac{40}{-10} = -4$
The solution is $(x, y) = (5, -4)$ .

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Solve the linear system using Cramer's Rule. Find the determinants $D$ , $D_x$ , and $D_y$ . The solution is $(x, y) = ($ $\frac{D_x}{D}$ $,$ $\frac{D_y}{D}$ $)$ .

STEP 1

STEP 2

Calculate the determinant $D$ of the coefficient matrix. The coefficient matrix is: $\begin{pmatrix} -3 & -1 \\ 2 & 4 \end{pmatrix}$ The determinant $D$ is calculated as: $D = (-3)(4) - (-1)(2)$

STEP 3

Compute the value of the determinant $D$ . $D = (-3)(4) - (-1)(2) = -12 - (-2) = -12 + 2 = -10$

STEP 4

Calculate the determinant $D_x$ by replacing the $x$ -coefficients with the constants from the right-hand side of the equations. The modified matrix is: $\begin{pmatrix} -11 & -1 \\ -6 & 4 \end{pmatrix}$ The determinant $D_x$ is calculated as: $D_x = (-11)(4) - (-1)(-6)$

STEP 5

Compute the value of the determinant $D_x$ . $D_x = (-11)(4) - (-1)(-6) = -44 - 6 = -50$

STEP 6

Calculate the determinant $D_y$ by replacing the $y$ -coefficients with the constants from the right-hand side of the equations. The modified matrix is: $\begin{pmatrix} -3 & -11 \\ 2 & -6 \end{pmatrix}$ The determinant $D_y$ is calculated as: $D_y = (-3)(-6) - (2)(-11)$

STEP 7

Compute the value of the determinant $D_y$ . $D_y = (-3)(-6) - (2)(-11) = 18 + 22 = 40$

STEP 8

Now that we have the determinants $D$ , $D_x$ , and $D_y$ , we can find the values of $x$ and $y$ using Cramer's Rule. $x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$

STEP 9

Calculate the value of $x$ . $x = \frac{D_x}{D} = \frac{-50}{-10} = 5$

STEP 10

Calculate the value of $y$ . $y = \frac{D_y}{D} = \frac{40}{-10} = -4$
The solution is $(x, y) = (5, -4)$ .

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Solve the linear system using Cramer's Rule. Find the determinants DDD, DxD_xDx​, and DyD_yDy​. The solution is (x,y)=((x, y) = ((x,y)=(DxD\frac{D_x}{D}DDx​​,, ,DyD\frac{D_y}{D}DDy​​))).

STEP 1

STEP 2

Calculate the determinant DDD of the coefficient matrix. The coefficient matrix is: (−3−124) \begin{pmatrix} -3 & -1 \\ 2 & 4 \end{pmatrix} (−32​−14​) The determinant DDD is calculated as: D=(−3)(4)−(−1)(2) D = (-3)(4) - (-1)(2) D=(−3)(4)−(−1)(2)

STEP 3

Compute the value of the determinant DDD. D=(−3)(4)−(−1)(2)=−12−(−2)=−12+2=−10 D = (-3)(4) - (-1)(2) = -12 - (-2) = -12 + 2 = -10 D=(−3)(4)−(−1)(2)=−12−(−2)=−12+2=−10

STEP 4

STEP 5

Compute the value of the determinant DxD_xDx​. Dx=(−11)(4)−(−1)(−6)=−44−6=−50 D_x = (-11)(4) - (-1)(-6) = -44 - 6 = -50 Dx​=(−11)(4)−(−1)(−6)=−44−6=−50

STEP 6

STEP 7

Compute the value of the determinant DyD_yDy​. Dy=(−3)(−6)−(2)(−11)=18+22=40 D_y = (-3)(-6) - (2)(-11) = 18 + 22 = 40 Dy​=(−3)(−6)−(2)(−11)=18+22=40

STEP 8

Now that we have the determinants DDD, DxD_xDx​, and DyD_yDy​, we can find the values of xxx and yyy using Cramer's Rule. x=DxD,y=DyD x = \frac{D_x}{D}, \quad y = \frac{D_y}{D} x=DDx​​,y=DDy​​

STEP 9

Calculate the value of xxx. x=DxD=−50−10=5 x = \frac{D_x}{D} = \frac{-50}{-10} = 5 x=DDx​​=−10−50​=5

STEP 10

Calculate the value of yyy. y=DyD=40−10=−4 y = \frac{D_y}{D} = \frac{40}{-10} = -4 y=DDy​​=−1040​=−4The solution is (x,y)=(5,−4)(x, y) = (5, -4)(x,y)=(5,−4).

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Solve the linear system using Cramer's Rule. Find the determinants $D$ , $D_x$ , and $D_y$ . The solution is $(x, y) = ($ $\frac{D_x}{D}$ $,$ $\frac{D_y}{D}$ $)$ .

Calculate the determinant $D$ of the coefficient matrix. The coefficient matrix is: $\begin{pmatrix} -3 & -1 \\ 2 & 4 \end{pmatrix}$ The determinant $D$ is calculated as: $D = (-3)(4) - (-1)(2)$

Compute the value of the determinant $D$ . $D = (-3)(4) - (-1)(2) = -12 - (-2) = -12 + 2 = -10$

Compute the value of the determinant $D_x$ . $D_x = (-11)(4) - (-1)(-6) = -44 - 6 = -50$

Compute the value of the determinant $D_y$ . $D_y = (-3)(-6) - (2)(-11) = 18 + 22 = 40$

Now that we have the determinants $D$ , $D_x$ , and $D_y$ , we can find the values of $x$ and $y$ using Cramer's Rule. $x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$

Calculate the value of $x$ . $x = \frac{D_x}{D} = \frac{-50}{-10} = 5$

Calculate the value of $y$ . $y = \frac{D_y}{D} = \frac{40}{-10} = -4$
The solution is $(x, y) = (5, -4)$ .