Problem 48 - A Graphical Approach to Precalculus with Limits

Question

Solve each system analytically. If the equations are dependent, write the solution set in terms of the variable \(z\). $$\begin{aligned} x+2 y+z &=0 \\ 3 x+2 y-z &=4 \\ -x+2 y+3 z &=-4 \end{aligned}$$

Step-by-Step Solution

Answer

The solution set is \( \{ (x, y, z) : x = 2 + z, y = -1 - z, z = z \} \).

Step 1: Write the System of Equations

The given system of linear equations is as follows:1. \( x + 2y + z = 0 \)2. \( 3x + 2y - z = 4 \)3. \( -x + 2y + 3z = -4 \)

Step 2: Eliminate One Variable

To simplify the system, let's eliminate the variable \( z \). First, add equation 1 and equation 2:\[ (x + 2y + z) + (3x + 2y - z) = 0 + 4 \]Simplifying gives:\[ 4x + 4y = 4 \]Divide throughout by 4:\[ x + y = 1 \] Now eliminate \( z \) using equations 2 and 3. Add equation 2 and equation 3:\[ (3x + 2y - z) + (-x + 2y + 3z) = 4 + (-4) \]Simplifying gives:\[ 2x + 4y + 2z = 0 \]Divide throughout by 2:\[ x + 2y + z = 0 \]Notice equation 1 and this result are identical, which means the system is dependent.

Step 3: Express the Solution in Terms of One Variable

Since the system is dependent, express \( x \) and \( y \) in terms of \( z \) using the simplified equations. From equation: \[ x + y = 1 \] Express \( x \) in terms of \( y \): \[ x = 1 - y \]Substitute this into equation 1, \( x + 2y + z = 0 \):\[ (1 - y) + 2y + z = 0 \]\[ 1 + y + z = 0 \]Thus, express \( y \) in terms of \( z \):\[ y = -1 - z \]Substitute back to find \( x \):\[ x = 1 - (-1 - z) = 2 + z \]

Step 4: Solution Set

The solution set in terms of \( z \) (where \( z \) is a parameter) is:\[ \{ (x, y, z) : x = 2 + z, y = -1 - z, z = z \} \]

Key Concepts

Dependent Systems

Dependent systems are systems of linear equations in which all equations describe the same plane or line in the solution space. This means they have infinitely many solutions, as opposed to having a unique solution or no solution at all. In a dependent system, each equation can be expressed as a linear combination of the others.

In the given problem, the system of equations appears to be dependent because, after manipulating the equations, we end up with identical equations. This indicates that the equations describe the same geometric object. As a result, any solution that satisfies one equation will satisfy the others, leading to an entire line or plane of solutions, not a single point.

Recognizing a dependent system is crucial in solving systems analytically because it helps in understanding the nature of solutions and dictates how to appropriately express the solution set.

Elimination Method

The elimination method is a systematic approach to solving systems of equations by removing one variable at a time. This is often achieved by adding or subtracting equations to cancel out one of the variables, which simplifies the system to fewer equations with fewer unknowns.

In our problem, we used the elimination method to remove the variable \( z \) from the equations. By manipulating the given equations, we formed new equations where \( z \) was eliminated, allowing us to observe dependencies among \( x \) and \( y \).

• First, we combined equations to eliminate \( z \) and simplify the system.
• By comparing the resulting equations, we determined the dependency.

This method is effective because it reduces the complexity of the problem, making it easier to interpret and solve the system. It is especially useful in identifying dependent systems, as demonstrated in this case.

Solution Set

The solution set of a system of equations is the set of all values that satisfy every equation in the system. For a dependent system, this set is infinite, often described using a parameter that represents one of the variables.

In this exercise, because our system is dependent, the solution cannot be expressed as a single point. Instead, we express the solution set in terms of a free variable \( z \).

• Solutions are given as a set of expressions for \( x \) and \( y \) in terms of \( z \).
• By allowing \( z \) to take any real value, we describe all possible solutions.

The solution set presented in the problem is: \[\{ (x, y, z) : x = 2 + z, y = -1 - z, z = z \}\]This notation effectively conveys that there are infinitely many solutions forming a line in three-dimensional space, emphasized by making \( z \) a parameter.

Linear Equations

Linear equations are algebraic equations in which each term is either a constant or the product of a constant and a single variable. These equations graph as straight lines when plotted on a coordinate plane.

In systems of linear equations, we explore interactions between multiple lines to find common intersection points, known as solutions. This is represented as:

• Graphically, these solutions are where lines intersect.
• Algebraically, solving them helps us find these intersection points.

The equations in our system are linear because they each describe a plane in three-dimensional space. Observing how these "planes" intersect provides the solutions to the system, as demonstrated by finding the relation between the variables and expressing one or more in terms of others, where possible.

Understanding linear equations is foundational for solving more complicated forms of algebraic equations and is a critical skill in mathematics.