Problem 28 - Precalculus with Limits

Question

Solving a System of Linear Equations In Exercises \(25 - 46\) , solve the system of linear equations and check any solutions algebraically. $$\left\\{ \begin{aligned} 2 x + 4 y + z = & 1 \\ x - 2 y - 3 z = & 2 \\ x + y - z = & \- 1 \end{aligned} \right.$$

Step-by-Step Solution

Answer

The system of equations has no solutions.

Step 1: Simplifying the system

Let's start by subtracting the third equation from the second. This results in a new system of equations as follows: \[\begin{aligned} 2x + 4y + z &= 1 \ 0x – 3y – 2z &= 3 \ x + y - z &= -1 \end{aligned}\] This kills the x-term in the second equation, simplifying the system.

Step 2: Solving for one variable

Now, let's multiply the second equation by 2 and add it to the first equation. This gives us a new system: \[\begin{aligned} 2x + 0y + 0z &= 7 \ 0x – 3y – 2z &= 3 \ x + y - z &= -1 \end{aligned}\] From the first equation, we then find that \( x = \frac{7}{2} \).

Step 3: Solving for the other two variables

Substitute \( x = \frac{7}{2} \) into the third and second equations to get: \[\begin{aligned} \frac{7}{2} + y - z &= -1 \ - 3y - 2z &= 3 \end{aligned}\]From the first equation, we get \( y - z = -\frac{9}{2} \). Now multiply the first of these results by 3 and the second by 2 and add them together: \(0y + 0z = 13\). This is not possible, so the given system of equations has no solutions.

Step 4: Checking the solution

Since the system does not have solutions, it is impossible to substitute the non-existent solutions back into the original system to check also for non-existent solutions. Therefore, the check is not necessary.

Key Concepts

algebraic solutions

Algebraic solutions are methods used to find the values of variables that satisfy a system of equations. In the case of linear systems, like the one in the given exercise, these solutions involve using different algebraic techniques to simplify the equations and identify solutions.

There are several ways to find algebraic solutions, including:

• Substitution - Solving one equation for one variable and substituting this expression into the other equations.
• Elimination - Adding or subtracting equations to eliminate one of the variables.
• Matrix methods - Using matrices and row reductions to find solutions.

In our example, the algebraic solution was approached using the technique of elimination to progressively simplify the system. This process was strategic in that it aimed to isolate variables or simplify relationships between them until a conclusion about the system's consistency could be reached.

inconsistent systems

An inconsistent system of linear equations is a system that has no solutions. This occurs when the equations represent parallel planes or lines that never intersect. When dealing with such systems, they often end up with a contradictory result such as a statement that is always false (e.g., \(0 = 5\)).
In the provided exercise, after attempting to simplify and reduce the system using algebraic techniques, we were left with the equation \(0y + 0z = 13\). This equation suggests a logical inconsistency, as it translates to \(0 = 13\), which is impossible and therefore indicates that the original system of equations has no solutions. Recognizing inconsistency in a system is crucial as it saves time by not pursuing further calculations unnecessarily. It allows one to understand the nature of the relationship among the equations involved.

variables substitution

Variables substitution is a method employed to simplify systems of equations. It involves solving one of the equations for one variable in terms of the other variables and then substituting this expression into other equations, which reduces the number of variables.

This method was used in our original exercise when substituting \(x = \frac{7}{2}\) back into the other equations. By doing this, we could eliminate one variable completely and focus on solving for the remaining two. Substitution is particularly useful when one of the equations is already simple enough to express one of the variables directly in terms of others. When combined with other techniques, like elimination, it becomes a powerful tool to simplify and solve complex systems of equations.