Problem 33 - A Graphical Approach to Precalculus with Limits

Question

Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. $$\begin{array}{r} x+y=-3 \\ 2 x-5 y=-6 \end{array}$$

Step-by-Step Solution

Answer

The solution is the ordered pair \((-3, 0)\).

Step 1: Write the System as an Augmented Matrix

Convert the given system of equations into an augmented matrix. The system \(x + y = -3\) and \(2x - 5y = -6\) is written as: \[\begin{bmatrix} 1 & 1 & | & -3 \ 2 & -5 & | & -6 \end{bmatrix}\]

Step 2: Eliminate the First Column Below the First Row

To zero out the element in the second row, first column, use the first row. Subtract 2 times the first row from the second row:\[R_2 = R_2 - 2R_1\]\[\begin{bmatrix} 1 & 1 & | & -3 \ 0 & -7 & | & 0 \end{bmatrix}\]

Step 3: Solve for the Second Variable

The second row now simplifies to \( -7y = 0 \). Solve for \(y\):\[ y = \frac{0}{-7} = 0 \]

Step 4: Back-Substitute to Find the First Variable

Substitute \(y = 0\) back into the first equation to find \(x\):\[ x + 0 = -3 \]\[ x = -3 \]

Step 5: Write the Solution as an Ordered Pair

The solution to the system is the ordered pair \( (x, y) = (-3, 0) \).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving a system of equations means finding the values of the variables that make all the equations true at the same time.
A simple example is a system like this:

• Equation 1: \( x + y = -3 \)
• Equation 2: \( 2x - 5y = -6 \)

Here, the variables are \( x \) and \( y \). By solving this system, we determine the specific values for \( x \) and \( y \) that satisfy both equations together. Different methods exist for solving systems, such as graphing, substitution, and elimination. In this case, we'll utilize an augmented matrix and row operations to find the solution.

Row Operations

Row operations are actions performed on the rows of an augmented matrix to help find the solution to a system of equations. There are three main types of row operations:

• Swapping two rows
• Multiplying a row by a nonzero constant
• Adding or subtracting a multiple of one row to another row

These operations are applied with the goal of simplifying the matrix to a form from which the solutions can be easily identified. A primary objective is to create zeros below the main diagonal to move towards the goal of obtaining row-echelon form. In the provided example, we used row operations to zero out the element below the first column's first row. Specifically, this involved subtracting two times the first row from the second row.

Solution of a System

The solution of a system is a set of values for the variables that satisfy all equations in the system simultaneously. In interpreting the solution from an augmented matrix, our goal is to simplify the system into easily solvable steps.
In our specific example, after using row operations, the matrix is simplified to reach a point where we can directly solve for one variable. This process yielded the second equation, \( -7y = 0 \). From here, we easily derived that \( y = 0 \). Subsequently, using back-substitution, we deduced the value for the remaining variable \( x \). Thus, the solution of this system is \( (x, y) = (-3, 0) \), telling us that when \( x = -3 \) and \( y = 0 \), both original equations are satisfied.

Back-Substitution

Back-substitution is the process of finding the values of variables in a system of equations after you have simplified the equations into a triangular form, generally through row operations or elimination methods.
Once the simpler equations have been determined, such as \( -7y = 0 \) in our example, you can substitute these known values into earlier equations to solve for other variables.
In essence, it is working backward from the simplest equation to the more complex ones. In our example, after finding \( y = 0 \), we substituted this back into the first equation, \( x + y = -3 \), to arrive at the solution \( x = -3 \). This step confirms the complete set of solutions and ensures the system of equations is fully satisfied with these variable values.