Problem 26 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{array}{rr} x+2 y+z= & 0 \\ 3 x+2 y-z= & 4 \\ -x+2 y+3 z= & -4 \end{array} $$

Step-by-Step Solution

Answer

The system has a parametric solution: \( (1-t, t, -1-t) \), where \( t \) is any real number.

Step 1: Write Down the System of Equations

We have the system of equations given by:1. \( x + 2y + z = 0 \)2. \( 3x + 2y - z = 4 \)3. \( -x + 2y + 3z = -4 \)

Step 2: Eliminate z from Equations 1 and 2

Add equation 1 and 2 to eliminate \( z \):\( (x + 2y + z) + (3x + 2y - z) = 0 + 4 \)This simplifies to:\( 4x + 4y = 4 \)Or simplified further:\( x + y = 1 \). Let this be equation 4.

Step 3: Eliminate z from Equations 1 and 3

Add equation 1 and 3 to eliminate \( z \):\( (x + 2y + z) + (-x + 2y + 3z) = 0 - 4 \)This simplifies to:\( 4y + 4z = -4 \)Or simplified:\( y + z = -1 \). Let this be equation 5.

Step 4: Solve for Variables x and y

From step 2 (equation 4), we have \( x + y = 1 \).We express \( x \) as \( x = 1 - y \).

Step 5: Solve for Variable z

Use step 3 (equation 5) to solve for \( z \):From \( y + z = -1 \), express \( z \) as \( z = -1 - y \).

Step 6: Find x, y, and z Solutions

Substitute \( y \) and solve each variable using parametric solution:From equation 4:- \( x = 1 - y \)From equation 5:- \( z = -1 - y \)Assume \( y = t \) (a parameter), then:- \( x = 1 - t \)- \( z = -1 - t \)Thus solutions are: \( (1-t, t, -1-t) \)

Step 7: Verify the Parametric Solution

Insert the parametric solution into each original equation and check if they hold true:- Equation 1: \((1-t) + 2t + (-1-t) = 0\) holds.- Equation 2: \(3(1-t) + 2t - (-1-t) = 4\) holds.- Equation 3: \(-(1-t) + 2t + 3(-1-t) = -4\) holds.Therefore, the system is consistent and has a solution.

Key Concepts

Elimination Method

In solving systems of equations, the elimination method is a powerful technique that allows us to eliminate one of the variables to simplify the problem.
Let's take the system of equations:

• \( x + 2y + z = 0 \)
• \( 3x + 2y - z = 4 \)
• \( -x + 2y + 3z = -4 \)

The goal is to merge two equations to cancel out one of the variables.
In this exercise, the first step is to add equations 1 and 2 to eliminate the variable \( z \):
- \( (x + 2y + z) + (3x + 2y - z) = 0 + 4 \)
This simplifies to \( 4x + 4y = 4 \), or \( x + y = 1 \).The process continues by eliminating \( z \) from another combination:
Add equations 1 and 3:

• \( (x + 2y + z) + (-x + 2y + 3z) = 0 - 4 \)

This results in \( 4y + 4z = -4 \), simplifying to \( y + z = -1 \).
Now, we have a simpler system we can solve for \( x \) and \( y \).
The elimination method effectively reduces the complexity of the original system.

Parametric Solutions

A parametric solution involves expressing variables in terms of a parameter, often denoted by \( t \).
In the context of systems of equations, parametric solutions are useful when there are infinitely many solutions. In this exercise, after using the elimination method, we found:

• \( x + y = 1 \)
• \( y + z = -1 \)

To solve this using parameters, assume \( y = t \). Then, each variable can be expressed as:

• \( x = 1 - t \)
• \( z = -1 - t \)

By introducing the parameter \( t \), we successfully represent the entire solution set as \( (1-t, t, -1-t) \).
This indicates that any value of \( t \) offers a different solution, showcasing the system's infinite solutions.

Consistency of Systems

The consistency of a system of equations refers to whether there is at least one solution. A consistent system can have one solution, infinitely many solutions, or no solution at all. In our exercise, after finding the parametric solutions, we insert the solutions back into the original equations to confirm consistency:

• Substituting \( t \) into the equations checked whether both \( x, y, \) and \( z \) satisfy all the equations.
• Equation 1, 2, and 3 all hold true for our parametric solution \( (1-t, t, -1-t) \).

This confirmation means the system is consistent, as each equation is satisfied by the solutions.
Thus, understanding consistency ensures the integrity of our solution.