Problem 21 - College Algebra

Question

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{l}{2 x-3 y=-9} \\ {5 x+4 y=58}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{96}{29}\) and \(y = \frac{151}{29}\).

Step 1: Write the system of equations as an augmented matrix

First, we write the given system of equations as an augmented matrix:\[\begin{bmatrix}2 & -3 & | & -9 \5 & 4 & | & 58\end{bmatrix}\]

Step 2: Perform row operations to form a row-echelon matrix

We want to form a row-echelon matrix. Let's start by eliminating the '5' from the second row by expressing row 2 as a linear combination:\[ R_2 = R_2 - \frac{5}{2}R_1 \]The operation results in: \[\begin{bmatrix}2 & -3 & | & -9 \0 & \frac{29}{2} & | & \frac{151}{2}\end{bmatrix}\]

Step 3: Obtain leading 1 in the second row

Our next step is to obtain a leading 1 in the second row. We can achieve this by multiplying the second row by \(\frac{2}{29}\):\[ R_2 = \frac{2}{29}R_2 \]Resulting in:\[\begin{bmatrix}2 & -3 & | & -9 \0 & 1 & | & \frac{151}{29}\end{bmatrix}\]

Step 4: Back-substitution to solve for y

With the matrix in row-echelon form, we start solving for \(y\) using the second equation:\[ y = \frac{151}{29} \]

Step 5: Solve for x using substitution

Substitute the value of \(y\) back into the first equation:\[ 2x - 3\left( \frac{151}{29} \right) = -9 \]Simplify and solve for \(x\):\[ 2x - \frac{453}{29} = -9 \]Convert \(-9\) to a fraction with the common denominator:\[ 2x = -9 + \frac{453}{29} \]Now multiply both sides by 29:\[ 58x = -261 + 453 \]\[ 58x = 192 \]Divide by 58:\[ x = \frac{192}{58} = \frac{96}{29} \]

Step 6: Conclusion: Solution to the System

The solution to the system is \(x = \frac{96}{29}\) and \(y = \frac{151}{29}\). Hence, we have found the required values of \(x\) and \(y\) using Gaussian elimination.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations that share common variables. In other words, the solution to the system is the set of values that satisfy all equations simultaneously. In our example, we have the system: \(2x - 3y = -9\) and \(5x + 4y = 58\). These two equations must hold true at the same time.

When solving a system of equations, the goal is to find the exact values of the variables involved. Systems can be classified as either having:

• One unique solution
• No solution (inconsistent system)
• Infinitely many solutions

Gaussian elimination is one of the techniques used to manage and solve such systems efficiently.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It contains the coefficients of the variables and the constants from the equations on the right side of an equal sign.

For the system \(2x - 3y = -9\) and \(5x + 4y = 58\), the augmented matrix is:\[\begin{bmatrix}2 & -3 & | & -9 \ 5 & 4 & | & 58\end{bmatrix}\]

The vertical line divides the coefficients and the constants, fashioning a unified structure for easier manipulation. It simplifies complex linear algebra processes by offering a clear visual of the operations needed. Transforming this matrix using Gaussian elimination can lead us toward finding a solution for the system.

Row-echelon Form

Row-echelon form is a version of a matrix where each leading term (non-zero entry) of a row is to the right of the leading term of the previous row. Moreover, all nonzero rows are above any rows of all zeros.

In our exercise, we transform the original augmented matrix into row-echelon form using row operations such as:

• Multiplying a row by a nonzero scalar
• Adding or subtracting the multiples of rows
• Swapping two rows

By applying these operations, the matrix was rearranged into:\[\begin{bmatrix}2 & -3 & | & -9 \ 0 & 1 & | & \frac{151}{29}\end{bmatrix}\]

Here, the zero below the leading 2 in the first column indicates the start of the row-echelon form, paving the way for easy backward substitution to find the variables.

Back-substitution

Back-substitution is a technique used after a matrix is reduced to row-echelon form. It's the process of solving for variable values starting from the last row that has been reduced, working upwards one row at a time.

In our problem, once the augmented matrix was in row-echelon form:\[\begin{bmatrix}2 & -3 & | & -9 \ 0 & 1 & | & \frac{151}{29}\end{bmatrix}\]we began solving by stating that \(y = \frac{151}{29}\) from the second row which has been simplified to a single equation in terms of \(y\).

After determining \(y\), this value was substituted into the first equation to solve for \(x\). Back-substitution is crucial because it capitalizes on the simpler form of the matrix where one variable can be easily isolated and solved, working its way through the system until all variables are determined.