Problem 22 - College Algebra

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{l}{6 x+2 y=-4} \\ {3 x+4 y=-17}\end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 1\) and \(y = -5\).

Step 1: Write the system in augmented matrix form

First, express the given system of equations as an augmented matrix. The augmented matrix represents the system:\[ \begin{bmatrix} 6 & 2 & | & -4 \ 3 & 4 & | & -17 \end{bmatrix} \]

Step 2: Perform row operations to form an upper triangular matrix

To form an upper triangular matrix, aim to have zeros below the diagonal. Start by making the element in the first column of the second row zero by replacing Row 2 with Row 2 minus half of Row 1:\[\begin{bmatrix} 6 & 2 & | & -4 \0 & 3 & | & -15 \end{bmatrix}\]

Step 3: Back substitution to find the value of y

Now that you have an upper triangular matrix, solve for \(y\) using the second row, which gives the equation:\[ 3y = -15 \]Divide both sides by 3 to find:\[ y = -5 \]

Step 4: Substitute y value to find x

Substitute \(y = -5\) into the first original equation:\[6x + 2(-5) = -4\]Which simplifies to:\[6x - 10 = -4\]Add 10 to both sides:\[6x = 6\]Divide both sides by 6 to solve for \(x\):\[x = 1\]

Step 5: Verify the solution

Substitute \(x = 1\) and \(y = -5\) back into the original equations to verify the solution. For the first equation:\[ 6(1) + 2(-5) = -4 \]This simplifies to \(-4 = -4\), which is true. For the second equation:\[ 3(1) + 4(-5) = -17 \]This simplifies to \(-17 = -17\), which is also true. The solution is verified.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In the given problem, the system consists of two linear equations with two variables, \(x\) and \(y\):

• \(6x + 2y = -4\)
• \(3x + 4y = -17\)

This type of system can be solved using various methods like substitution, elimination, or matrix methods such as Gaussian elimination. The goal is to find the values for the unknowns that satisfy all equations in the system simultaneously. In this case, Gaussian elimination helps to systematically simplify the equations to easily find the solutions.

Upper Triangular Matrix

An upper triangular matrix is a type of matrix where all the elements below the main diagonal are zero. This matrix form is useful because it simplifies the process of solving equations using techniques like back substitution.
To achieve this form in Gaussian elimination, you perform row operations that aim to create zeros beneath the diagonal. In the example, after replacing Row 2 with Row 2 minus half of Row 1, the matrix becomes:

• \(\begin{bmatrix} 6 & 2 & | & -4 \ 0 & 3 & | & -15 \end{bmatrix}\)

This transformation makes it easier to solve for one of the variables using the non-zero equations, reducing the complexity of the system step by step.

Back Substitution

Back substitution is a method used to solve a system of equations once it has been transformed into an upper triangular matrix. Starting with the bottom row of the matrix, you solve for one variable and then substitute back to find the others.
In the example provided, the upper triangular matrix leads us to the equation \(3y = -15\). Solving this, we find \(y = -5\).
Once \(y\) is known, substitute it back into the first equation (\(6x + 2y = -4\)) to find \(x\). This process unravels the solution from the bottom row upward, hence termed "back" substitution.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficient matrix with the constant terms vector, separated by a vertical line.
For the given system, the corresponding augmented matrix is:

• \(\begin{bmatrix} 6 & 2 & | & -4 \ 3 & 4 & | & -17 \end{bmatrix}\)

This format streamlines the process of performing row operations and applying methods like Gaussian elimination. By focusing on numerical coefficients and constants directly, it simplifies procedure and reduces the room for arithmetic errors. Understanding augmented matrices is crucial for effectively using matrix methods in solving systems of equations.