Problem 21 - College Algebra with Corequisite Support

Question

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

Step-by-Step Solution

Answer

The solution is \((x, y) = (6, 7)\).

Step 1: Set Up the Augmented Matrix

First, we need to write the system of equations as an augmented matrix. The system is: \( \begin{cases} 2x - 3y = -9 \ 5x + 4y = 58 \end{cases} \) becomes the augmented matrix: \[ \begin{bmatrix} 2 & -3 & | & -9 \ 5 & 4 & | & 58 \end{bmatrix} \]

Step 2: Perform Row Operations to Get a Leading 1 in the First Row

Divide the first row by 2 to make the coefficient of \( x \) in the first row a 1: \[ \begin{bmatrix} 1 & -1.5 & | & -4.5 \ 5 & 4 & | & 58 \end{bmatrix} \]

Step 3: Eliminate x from the Second Row

Use the first row to eliminate \( x \) from the second row. Subtract 5 times the first row from the second row: \[ 5(1) - 5 = 0 \] for the first column, \[ 5(-1.5) - 4 = -7.5 - 4 = -11.5 \] for the second column, \[ 5(-4.5) - 58 = -22.5 - 58 = -80.5 \] for the constants. The new matrix is: \[ \begin{bmatrix} 1 & -1.5 & | & -4.5 \ 0 & -11.5 & | & -80.5 \end{bmatrix} \]

Step 4: Solve for y in the Second Row

Isolate \( y \) in the second row by dividing the entire second row by -11.5: \[ 0 y + 1 y = 7 \] The updated matrix becomes: \[ \begin{bmatrix} 1 & -1.5 & | & -4.5 \ 0 & 1 & | & 7 \end{bmatrix} \]. So, \( y = 7 \).

Step 5: Back-Substitution to Solve for x

Substitute \( y = 7 \) back into the first row equation to solve for \( x \): \[ 1x - 1.5(7) = -4.5 \] \[ x - 10.5 = -4.5 \] \[ x = 6 \].

Step 6: State the Solution

The solution to the system of equations is the point where both equations intersect: \( x = 6 \) and \( y = 7 \).

Key Concepts

Augmented Matrix

In Gaussian elimination, the first step is transforming a system of equations into an augmented matrix. This matrix includes the coefficients and constant terms of each equation. For example, the system\[ \begin{align*} 2x - 3y &= -9 \ 5x + 4y &= 58 \end{align*} \]can be represented as:\[ \begin{bmatrix} 2 & -3 & | & -9 \ 5 & 4 & | & 58 \end{bmatrix} \].
This notation combines both the variable coefficients and the constant terms on the right-hand side of the equations. This approach streamlines the process of manipulating and solving equations through row operations. By focusing on the numbers alone, we can efficiently apply operations without continually rewriting variables.

Row Operations

To use Gaussian elimination, we perform specific row operations on our augmented matrix. These operations help simplify the matrix and ultimately solve the system:

• Swap Rows: Change the order of the rows if needed to simplify calculations.
• Multiply a Row by a Non-zero Constant: For instance, dividing a row by a constant can help simplify coefficients.
• Add or Subtract Rows: Combine rows by adding or subtracting them to eliminate variables from rows.

In our example, we divided the first row by 2 to make the coefficient of \(x\) a 1:\[ \begin{bmatrix} 1 & -1.5 & | & -4.5 \ 5 & 4 & | & 58 \end{bmatrix} \].We then subtracted 5 times the first row from the second row to eliminate \(x\) from the second equation, transforming it further.

System of Equations

A system of equations is a set of equations with multiple variables. In our example, two linear equations intersecting at a specific point form the system:\[ \begin{align*} 2x - 3y &= -9 \ 5x + 4y &= 58 \end{align*} \].
Solving systems of equations helps us find the values of the variables which satisfy all equations simultaneously.
By applying Gaussian elimination and reducing the system to simpler forms, solutions become evident. The point \((x, y)\) where both lines in a 2D graph intersect represents the solution. This juncture reveals the particular \(x\) and \(y\) values satisfying both equations.

Back-Substitution

The final step in Gaussian elimination is back-substitution. After transforming the system into an upper triangular or simpler form, we solve for variables starting with the last equation.
In our example, with the simplified equations\[ y = 7 \]from the second row and substituting back into the first equation
\[ x - 10.5 = -4.5 \],we solve for \(x\). These calculations provide the solution for each variable and ultimately solve the system:

• \(y = 7\)
• \(x = 6\)

This method ensures clarity and precision in deriving variable values, as each preceding variable depends on the solutions of previous ones.