Problem 25 - 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 \(z\). $$\begin{array}{r} x-2 y+3 z=6 \\ 2 x-y+2 z=5 \end{array}$$

Step-by-Step Solution

Answer

The solution set is \((x, y, z) = \left(\frac{4z + 4}{3}, \frac{4z - 7}{3}, z\right)\) for any real number \(z\).

Step 1: Analyze the System of Equations

We have a system of two equations: 1. \( x - 2y + 3z = 6 \) 2. \( 2x - y + 2z = 5 \) Notice there are three variables but only two equations, indicating that the system might have infinitely many solutions, likely dependent on a parameter (here, \( z \)).

Step 2: Express One Variable in Terms of Others

To express one variable in terms of z, we can solve the first equation for \( x \):\[ x = 6 + 2y - 3z \]

Step 3: Substitute into Second Equation

Substitute \( x = 6 + 2y - 3z \) into the second equation:\[ 2(6 + 2y - 3z) - y + 2z = 5 \]Simplifying, we get:\[ 12 + 4y - 6z - y + 2z = 5 \]\[ 3y - 4z = -7 \]

Step 4: Solve for y in terms of z

Now solve \(3y - 4z = -7\) for \( y \):\[ 3y = 4z - 7 \]\[ y = \frac{4z - 7}{3} \]

Step 5: Express x in Terms of z

Substitute \( y = \frac{4z - 7}{3} \) back into \( x = 6 + 2y - 3z \):\[ x = 6 + 2 \left( \frac{4z - 7}{3} \right) - 3z \]\[ x = 6 + \frac{8z - 14}{3} - 3z \]\[ x = \frac{18}{3} + \frac{8z - 14 - 9z}{3} \]\[ x = \frac{4z + 4}{3} \]

Step 6: Write the Solution Set

The system is dependent, and the solution set can be expressed in terms of \( z \):\[ \left( x, y, z \right) = \left( \frac{4z + 4}{3}, \frac{4z - 7}{3}, z \right) \]Where \( z \) is any real number.

Key Concepts

Dependent Systems

In the context of systems of equations, a dependent system is one where the equations involved are not independent of each other. This means that one equation can be derived from another by manipulation, such as multiplication or addition. In simpler terms, the equations are linked in such a way that they represent the same relationship between variables.

In the given exercise, we have two equations involving three variables:

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

Since we have more variables than equations, and the equations do not contradict each other, we suspect the system to be dependent. Such a system typically has infinitely many solutions because the variables depend on each other, often through a parameter. In this specific case, that parameter is \( z \).

Rather than finding fixed values for \( x \), \( y \), and \( z \), we express two of these variables in terms of the third, usually chosen arbitrarily as a parameter. This results in a solution set that creates a line in three-dimensional space.

Analytical Solutions

An analytical solution to a system of equations involves using algebraic methods to solve the equations exactly. Unlike numerical solutions, which approximate answers using iterative methods, analytical solutions aim to express variables explicitly in terms of constants or parameters.

For the given system of equations, finding an analytical solution involves eliminating variables and expressing each in terms of the parameter \( z \). Here's how it works:

• Initially, solve for one variable in a single equation. In this case, solve for \( x \) in terms of \( y \) and \( z \).
• Substitute this expression into the other equation to find another relationship between variables, allowing us to solve for \( y \) in terms of \( z \).
• Finally, using these expressions, write a generalized solution for \( x \) and \( y \) with respect to \( z \).

The solution process is systematic and hinges on the ability to manipulate the equations algebraically to express as many variables as possible in terms of a chosen parameter. This method provides a clear path to determining the solution set for such systems.

Parameters in Solutions

Parameters in equations serve as placeholders where any real number can be input, leading to infinite possible solutions. This occurs commonly in dependent systems. In our example, the parameter \( z \) allows the determination of \( x \) and \( y \) based on its value.

Each choice of \( z \) corresponds to a unique set of \( x \) and \( y \). Let's break it down further:

• We first find \( y \) in terms of \( z \) as \( y = \frac{4z - 7}{3} \).
• Next, we substitute this into the expression for \( x \), giving us \( x = \frac{4z + 4}{3} \).

Thus, the solution set can be written as \( (x, y, z) = \left( \frac{4z + 4}{3}, \frac{4z - 7}{3}, z \right) \).

In essence, a parameter like \( z \) provides a dynamic quality to solutions. This approach covers infinitely many cases, not just fixed point solutions, making it valuable for understanding a full range of possible outcomes in variable-dependent scenarios.