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Solve each system, using Cramer's rule when possible. $$\begin{aligned} &\frac{1}{4} x+\frac{2}{3} y=25\\\ &\frac{3}{5} x-\frac{1}{10} y=12 \end{aligned}$$

Short Answer

Expert verified
The solution is x ≈ 24.7 and y ≈ 141.

Step by step solution

01

Write the system of equations in standard form

First, rewrite the given equations in standard form (Ax + By = C):\(\frac{1}{4} x + \frac{2}{3} y = 25\)\(\frac{3}{5} x - \frac{1}{10} y = 12\)
02

Express the equations in matrix form

Identify the coefficients from both equations to form matrices:The coefficient matrix is:\[ A = \begin{bmatrix} \frac{1}{4} & \frac{2}{3} \ \frac{3}{5} & -\frac{1}{10} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{bmatrix} \]The constants matrix is:\[ C = \begin{bmatrix} 25 \ 12 \end{bmatrix} \]
03

Compute the determinant of the coefficient matrix

The determinant of matrix A (detA) is computed as follows:\( det(A) = \frac{1}{4}(-\frac{1}{10}) - \frac{2}{3}(\frac{3}{5}) \)\( det(A) = -\frac{1}{40} - \frac{6}{15} \)\( det(A) = -\frac{1}{40} - \frac{4}{10} \)\( det(A) = -\frac{1}{40} - \frac{16}{40} \)\( det(A) = -\frac{17}{40} \)
04

Compute the determinants of the modified matrices

To find the determinant of the matrix when replacing the x-column (Dx):\[ A_{Dx} = \begin{bmatrix} 25 & \frac{2}{3} \ 12 & -\frac{1}{10} \ \ \ \ \ \ \end{bmatrix} \]\( det(A_{Dx}) = 25(-\frac{1}{10}) - \frac{2}{3}(12) \)\( det(A_{Dx}) = -\frac{25}{10} - \frac{24}{3} \)\( det(A_{Dx}) = -2.5 - 8 \)\( det(A_{Dx}) = -10.5 \)To find the determinant of the matrix when replacing the y-column (Dy):\[ A_{Dy} = \begin{bmatrix} \frac{1}{4} & 25 \ \frac{3}{5} & 12 \ \ \ \ \ \ \ \ \end{bmatrix} \]\( det(A_{Dy}) = \frac{1}{4}(12) - 25(\frac{3}{5}) \)\( det(A_{Dy}) = \frac{12}{4} - \frac{75}{5} \)\( det(A_{Dy}) = 3 - 15 \)\( det(A_{Dy}) = -72 \)
05

Solve for x and y using Cramer's rule

Using Cramer's rule, we find x and y as follows:\( x = \frac{det(A_{Dx})}{det(A)} \)\( x = \frac{ - 10.5 }{ - \frac{17}{40} } \)\( x = - 10.5 \frac{40}{-17} \)\( x = \frac{- 420.0}{-17} \)\( x = 24.7(rounded) \)\( y = \frac{det(A_{Dy})}{det(A)} \)\( y = \frac{-72}{ - \frac{17}{40}} \)\( y = -72 \frac{40}{-17} \)\( y = \frac{- 2400}{-17.0} \)\( y = 141.0(rounded) \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Equations
A system of equations is a collection of two or more equations with the same set of variables. Solving these systems involves finding values for the variables that satisfy all equations simultaneously. For example, the system of equations given in the exercise is represented as:

\( \frac{1}{4} x + \frac{2}{3} y = 25 \)
\( \frac{3}{5} x - \frac{1}{10} y = 12 \).

There are various methods to solve these systems, including substitution, elimination, and matrix methods like Cramer's Rule. Cramer's Rule is particularly useful for solving linear systems with the same number of equations as unknowns.
Determinants
A determinant is a special number calculated from a square matrix. It provides important properties of the matrix, such as whether the matrix is invertible. For a 2x2 matrix \(A\) represented as \[ A = \begin{bmatrix} a & b \ c & d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{bmatrix} \]

the determinant is calculated as \( \text{det}(A) = ad - bc \).

In the exercise, the determinant of the coefficient matrix \(A\) is computed as follows:

\( \text{det}(A) = \frac{1}{4} \times (-\frac{1}{10}) - \frac{2}{3} \times \frac{3}{5} \ \)
\( \text{det}(A) = -\frac{1}{40} - \frac{6}{15} \ \)
\( \text{det}(A) = -\frac{1}{40} - \frac{16}{40} \ \)
\( \text{det}(A) = -\frac{17}{40} \).
Matrix Algebra
Matrix algebra involves the manipulation of matrices to solve systems of linear equations and other mathematical problems. In this context, matrices are rectangular arrangements of numbers, symbols, or expressions in rows and columns. Here’s how matrices come into play:

  • The coefficient matrix \( A \) for the system is: \[ A = \begin{bmatrix} \frac{1}{4} & \frac{2}{3} \ \frac{3}{5} & -\frac{1}{10} \ \ \ \ \ \end{bmatrix} \]
  • The constants matrix \( C \): \[ C = \begin{bmatrix} 25 \ 12 \ \ \ \ \end{bmatrix} \]
  • Modified matrices for Cramer's Rule (replacing columns of \( A \) with \( C \)): \[ A_{Dx} = \begin{bmatrix} 25 & \frac{2}{3} \ 12 & -\frac{1}{10} \ \ \ \ \end{bmatrix} \], \[ A_{Dy} = \begin{bmatrix} \frac{1}{4} & 25 \ \frac{3}{5} & 12 \ \ \ \ \end{bmatrix} \]


Matrix algebra simplifies the calculation of determinants, as shown in the solution steps.
Linear Algebra
Linear algebra is a field of mathematics focusing on vector spaces, and linear transformations between those spaces. It includes concepts like vectors, matrices, and systems of linear equations. These are the fundamental tools used in solving our given system using Cramer's rule.

Linear equations like \( \frac{1}{4} x + \frac{2}{3} y = 25 \) form the basis of systems analyzed in linear algebra.

Cramer's rule specifically leverages matrix algebra and determinants to find the unique solutions of such systems. In this exercise, both concepts ensure a structured approach from formulation to solution:
  • Solve for \(x\) using: \( x = \frac{ \text{det}(A_{Dx}) }{ \text{det}(A) } \)
  • Solve for \(y\) using: \( y = \frac{ \text{det}(A_{Dy}) }{ \text{det}(A) } \)


  • Finding values of \(x\) and \(y\) becomes systematic and automated using these rules, highlighting the power of linear algebra in solving real-world problems.

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Most popular questions from this chapter

On the Bayou A-Bear's Catering Service charges its customers according to the number of servings of each item that is supplied at the party. The table shows the number of servings of jambalaya, crawfish pic, filé gumbo, iced tea, and dessert for the last five customers, along with the total cost of each party. What amount does A-Bear's charge per serving of each item? Timind Write a system of five equations in five unknowns and use a graphing calculator to solve it.\begin{aligned} &\begin{array}{|l|c|c|c|c|c|c|} \hline & \text { Jambalaya } & \text { Crawfish Pie } & \text { Filé Gumbo } & \text { Iced Tea } & \text { Dessert } & \text { Cost } \\ \hline \text { Boudreaux } & 36 & 28 & 35 & 90 & 68 & \$ 344.35 \\ \text { Thibodeaux } & 37 & 19 & 56 & 84 & 75 & \$ 369.10 \\ \text { Fontenot } & 49 & 55 & 70 & 150 & 125 & 5588.90 \\ \text { Arceneaur } & 58 & 34 & 52 & 122 & 132 & 5529.50 \\ \text { Gautreaux } & 44 & 65 & 39 & 133 & 120 & 5521.65 \\ \hline \end{array}\\\ &v \end{aligned}

The following exercises investigate some of the properties of determinants. For these exercises let \(M=\left[\begin{array}{ll}3 & 2 \\ 5 & 4\end{array}\right]\) and \(N=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right]\). Prove that if \(A\) is any \(2 \times 2\) invertible matrix, then the determinant of \(A^{-1}\) is the reciprocal of the determinant of \(A\)

Let \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right],\) and \(C=\left[\begin{array}{ll}c_{11} & c_{12} \\\ c_{21} & c_{22}\end{array}\right]\) Determine whether each of the following statements is true, and explain your answer. For any real numbers \(s\) and \(t, s A+t A=(s+t) A\).

Solve each problem. Find all matrices of the form \(A=\left[\begin{array}{ll}a & a \\ 0 & c\end{array}\right]\) such that \(A^{2}=I\).

Solve each system of equations by using } A^{-1} \text { if possible.}$$ $$\begin{aligned} x-y+z &=5 \\ 2 x-y+3 z &=1 \\ y+z &=-9 \end{aligned}$$

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