Problem 24 - College Algebra

Question

Solve the system by Gaussian elimination. \(\begin{aligned}-4 x-3 y &=-2 \\ 3 x-5 y &=-13 \end{aligned}\)

Step-by-Step Solution

Answer

The solution is \(x = -1\) and \(y = 2\).

Step 1 - Write the Augmented Matrix

First, we need to write the system of equations as an augmented matrix. The given system is: \[-4x - 3y = -2\] \[3x - 5y = -13\] The augmented matrix for this system is: \[\begin{bmatrix} -4 & -3 & | & -2 \ 3 & -5 & | & -13 \end{bmatrix}\]

Step 2 - Swap Rows for Easier Elimination

To make calculations simpler, we'll swap the rows to start elimination with a positive coefficient for \(x\). After swapping the rows, the matrix looks like: \[\begin{bmatrix} 3 & -5 & | & -13 \ -4 & -3 & | & -2 \end{bmatrix}\]

Step 3 - Eliminate \(x\) from the Second Row

We need to eliminate the \(x\)-term from the second row. To do this, multiply the first row by 4 and the second row by 3, then add them and replace the second row: \[4 \times \begin{bmatrix} 3 & -5 & | & -13 \end{bmatrix} = \begin{bmatrix} 12 & -20 & | & -52 \end{bmatrix}\] \[3 \times \begin{bmatrix} -4 & -3 & | & -2 \end{bmatrix} = \begin{bmatrix} -12 & -9 & | & -6 \end{bmatrix}\] \[\text{Add them: } \begin{bmatrix} 0 & -29 & | & -58 \end{bmatrix}\] The new matrix is: \[\begin{bmatrix} 3 & -5 & | & -13 \ 0 & -29 & | & -58 \end{bmatrix}\]

Step 4 - Solve for \(y\)

Now, solve the second row for \(y\). The equation is: \[ -29y = -58 \] Divide both sides by -29: \[ y = 2 \]

Step 5 - Back Substitute to Find \(x\)

Substitute \(y = 2\) into the first row's equation to find \(x\): \[3x - 5(2) = -13\] This simplifies to: \[3x - 10 = -13\] Add 10 to both sides: \[3x = -3\] Divide by 3: \[x = -1\]

Key Concepts

Augmented Matrix

An augmented matrix is a powerful representation used to solve systems of linear equations. It consolidates the coefficients of variables and the solutions found in linear equations into a single matrix format. This makes solving systems more organized and algebraically efficient.
To construct an augmented matrix from a linear equations system, arrange the coefficients of the variables in the equations, and place the constants on the other side of the equation as an additional column.
For example, consider the system:

• \(-4x - 3y = -2\)
• \(3x - 5y = -13\)

The augmented matrix representing this system is:\[\begin{bmatrix} -4 & -3 & | & -2 \3 & -5 & | & -13 \end{bmatrix}\]The vertical line separates the coefficients from the constants, signifying the equals sign in the equations. This matrix now becomes the foundation for applying Gaussian elimination.

Elimination Method

The elimination method, often referred to as Gaussian elimination, is a technique to solve a system of linear equations. The objective is to transform the system into a simpler form that can be solved easily through back substitution.
You can do this by performing row operations on the augmented matrix: swapping rows, multiplying a row by a constant, or adding rows. These operations help you eliminate variables step by step.
In the exercise given, we started by swapping the rows to make calculations more straightforward. By making sure the leading term (coefficient of the first variable) is positive, we facilitate easier calculations:\[\begin{bmatrix} 3 & -5 & | & -13 \-4 & -3 & | & -2 \end{bmatrix}\]Next, eliminate the variable in the lower rows by combining rows strategically. For example, multiply rows if necessary to align variables and add or subtract them to eliminate one of the variables from certain rows. This provides a more manageable system and allows for clear solutions to emerge.

Back Substitution

Back substitution is the final phase in the Gaussian elimination process. Once you have your matrix in an upper triangular form, where all elements below the main diagonal are zero, you solve the equations from the last row upwards.
Starting with the simplest equation, calculate the values of unknowns using previously obtained results.
In our example, once the matrix has been reduced to:\[\begin{bmatrix}3 & -5 & | & -13 \0 & -29 & | & -58\end{bmatrix}\]You begin solving from the bottom upwards:

• Solve \(-29y = -58\) to find \(y = 2\).
• Substitute \(y = 2\) back into the first equation, \(3x - 5(2) = -13\), to find \(x = -1\).

By solving from the simplest form up, you consolidate your results progressively, giving you the solution set for the variables in the equations.

Linear Equations System

A linear equations system is a collection of one or more equations involving the same set of variables. These equations can be represented graphically as straight lines in a coordinate plane if there are two variables.
The solution to such a system is a set of values for the variables that satisfy all equations simultaneously, often where the equations' graphs intersect.
The system examined in our exercise consisted of:

• \(-4x - 3y = -2\)
• \(3x - 5y = -13\)

This can be pictorially seen as two straight lines crossing each other, where the point of intersection provides the solution to both equations. By converting these equations into an augmented matrix and employing the Gaussian elimination technique, you simplify the process of finding this intersection. This allows for quick and effective solving without graphing.