/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Solve the system of equations us... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the system of equations using Gaussian elimination or Gauss-Jordan elimination. Use a graphing calculator to check your answer. $$\begin{aligned} &5 x-3 y=-2\\\ &4 x+2 y=5 \end{aligned}$$

Short Answer

Expert verified
\(x = 1\) and \(y = \frac{3}{2}\)

Step by step solution

01

Write the augmented matrix

Convert the system of equations into an augmented matrix. \(\begin{bmatrix} 5 & -3 & | & -2 \ 4 & 2 & | & 5 \end{bmatrix}\)
02

Apply the row operations

Use row operations to transform the matrix into row-echelon form. First, make the element in the first row, first column a 1 by dividing the entire first row by 5. \(\begin{bmatrix} 1 & -\frac{3}{5} & | & -\frac{2}{5} \ 4 & 2 & | & 5 \end{bmatrix}\)
03

Eliminate the first column below the first row

Subtract 4 times the first row from the second row to eliminate the first column below the first row.\(\begin{bmatrix} 1 & -\frac{3}{5} & | & -\frac{2}{5} \ 0 & \frac{22}{5} & | & \frac{33}{5} \end{bmatrix}\)
04

Normalize the second row

Divide the second row by \(\frac{22}{5}\) to have 1 in the second row, second column.\(\begin{bmatrix} 1 & -\frac{3}{5} & | & -\frac{2}{5} \ 0 & 1 & | & \frac{3}{2} \end{bmatrix}\)
05

Eliminate the second column above the second row

Add \(\frac{3}{5}\) times the second row to the first row to eliminate the second column above the second row.\(\begin{bmatrix} 1 & 0 & | & 1 \ 0 & 1 & | & \frac{3}{2} \end{bmatrix}\)
06

Solution from row-echelon form

Extract the solutions from the resulting matrix. \(x = 1\) and \(y = \frac{3}{2}\)
07

Verify the solution

Use a graphing calculator to substitute \(x = 1\) and \(y = \frac{3}{2}\) into the original equations to verify the solution.

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Key Concepts

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

Gauss-Jordan elimination
Gauss-Jordan elimination is a straightforward method to solve systems of linear equations. This process involves transforming a system's augmented matrix into reduced row-echelon form. Unlike Gaussian elimination which stops at row-echelon form, Gauss-Jordan elimination simplifies further to make back-substitution unnecessary. By the end, each variable will be isolated in its own row, making it easy to read off the solutions directly. This method efficiently solves for variables in a step-by-step manner using basic row operations.
Systems of equations
A system of equations is a set of equations with multiple variables. The goal is to find values for these variables that satisfy all the equations simultaneously. For example, the given system of equations is:
\(5x - 3y = -2\)
\(4x + 2y = 5\).
Solving such a system can be complex if done by substitution or elimination manually, especially for larger systems. Methods like Gaussian and Gauss-Jordan elimination make it systematic and efficient to find the solution. In this instance, the solution to the system is \(x = 1\) and \(y = \frac{3}{2}\).
Augmented matrix
An augmented matrix combines the coefficients and constants of a system of equations into a single matrix. This matrix makes it easier to perform row operations. For our system:
\(5 x - 3 y = -2\)
\(4 x + 2 y = 5\),
the augmented matrix looks like this:
\[ \begin{matrix} 5 & -3 & | & -2 & \ 4 & 2 & | & 5 \ \ \ end{bmatrix} \].
This format helps centralize the information and allows us to focus on manipulating the rows to simplify the system.
Row operations
Row operations are essential tools in methods like Gauss-Jordan elimination. These operations include:
  • Row swapping: Changing the positions of two rows.

  • Row multiplication: Multiplying all elements of a row by a non-zero scalar.

  • Row addition: Adding or subtracting multiples of one row to another row.


These operations help transform the augmented matrix step by step until it reaches reduced row-echelon form. In our solution, row operations simplified the matrix and eventually led us to find \(x = 1\) and \(y = \frac{3}{2}\). Row operations allow us to handle even complex systems systematically and efficiently.

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Most popular questions from this chapter

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