Problem 35 - 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 \(z\). $$\begin{aligned} &x-y+z=-6\\\ &x+y+3 z=10 \end{aligned}$$

Step-by-Step Solution

Answer

Solution set: \((x, y, z) = (2 - 2z, 8 - z, z)\).

Step 1: Label the Equations

Label the given equations for clarity: \( (1) : x - y + z = -6 \) and \( (2) : x + y + 3z = 10 \).

Step 2: Eliminate Variable

Add equations (1) and (2) to eliminate \( y \):\[ \begin{align*} (1) + (2) : &\ (x - y + z) + (x + y + 3z) = -6 + 10 \&\ 2x + 4z = 4 \&\ x + 2z = 2 \ \end{align*} \]},

Step 3: Express x in terms of z

From the equation \( x + 2z = 2 \), solve for \( x \): \[ x = 2 - 2z \].

Step 4: Solve for y in terms of z

Substitute \( x = 2 - 2z \) into equation (1):\[ \begin{align*} (2 - 2z) - y + z &= -6 \ -y &= -6 - 2 + z \ -y &= -8 + z \ y &= 8 - z \end{align*} \]

Step 5: Formulate the Solution Set

With \( x = 2 - 2z \) and \( y = 8 - z \), the solution set in terms of \( z \) is: \[ (x, y, z) = (2 - 2z, 8 - z, z) \].

Key Concepts

Dependent Equations

In the context of systems of equations, dependent equations are equations that describe the same line or plane. This happens when one equation can be derived from another by multiplication or addition. When this occurs, both equations provide the same solution information, resulting in infinitely many solutions. In other words, the equations do not provide unique information to determine distinct variable values.
When solving a system with dependent equations, you'll often be able to express the solutions in terms of a free variable, as seen in the original exercise where the solution was expressed in terms of the variable \( z \). This indicates that as \( z \) varies, \( x \) and \( y \) vary accordingly, showing the dependency of the equations on one another.

Solution Set

The solution set of a system of equations refers to all possible values of the variables that satisfy all equations in the system. In some cases, like with dependent equations, the solution set is infinite. This means an infinite number of combinations of the variables exist that solve the system.
In the provided exercise, the solution set is expressed in terms of \( z \), resulting in a general solution given by \( (x, y, z) = (2 - 2z, 8 - z, z) \). This notation represents every point that satisfies the original equations by varying \( z \). This form reveals the dependency on \( z \) as a free parameter, which allows for the description of a line or plane where all points satisfy the system.

Elimination Method

The elimination method is a common technique for solving systems of equations. The idea is to remove one variable to make the system easier to solve. This is achieved by adding or subtracting equations strategically to cancel a variable.
In the exercise provided, equations were added to eliminate the \( y \) variable:

• Add Equation 1: \( x - y + z = -6 \)
• and Equation 2: \( x + y + 3z = 10 \)

By adding these, the \( y \) terms cancel each other out, simplifying the system to a single equation in terms of \( x \) and \( z \): \( x + 2z = 2 \). This reduced equation makes it easier to solve for one variable and identify the relationship between the remaining variables.

Variable Substitution

Variable substitution is a method used to simplify solving equations by expressing one or more variables in terms of others. After reducing a system using elimination, substitution allows for further simplification.
In the solution provided, once the equation \( x + 2z = 2 \) was solved for \( x \) yielding \( x = 2 - 2z \), substitution was used to express \( y \) in terms of \( z \). The value for \( x \) derived from substitution was plugged back into one of the original equations, allowing for the isolation of \( y \):

• From \( (2 - 2z) - y + z = -6 \), solve for \( y \) to find \( y = 8 - z \).

This step-by-step approach makes solving more manageable by systematically reducing the complexity of the equations. The final solution expressed in terms of \( z \) effectively utilizes substitution to outline the relationship among the variables.