Problem 44 - 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 \(z\). $$\begin{array}{r} 2 x+y+3 z=4 \\ -3 x-y-4 z=5 \\ x+y+2 z=0 \end{array}$$

Step-by-Step Solution

Answer

The solution set is \((x, y, z) = \left(-\frac{5}{2} - z, \frac{5}{2} - z, z\right)\).

Step 1: Label the Equations

For clarity, let's label the given system of equations as follows: \( (1) \) \( 2x + y + 3z = 4 \), \( (2) \) \( -3x - y - 4z = 5 \), and \( (3) \) \( x + y + 2z = 0 \).

Step 2: Eliminate y from Equations (1) and (3)

Add equations \( (1) \) and \( (3) \):\( (2x + y + 3z) + (x + y + 2z) = 4 + 0 \), which simplifies to \( 3x + 2y + 5z = 4 \). We now consider this our new equation \( (4) \).

Step 3: Eliminate y from Equations (2) and (3)

Add equations \( (2) \) and \( (3) \):\( (-3x - y - 4z) + (x + y + 2z) = 5 + 0 \), which simplifies to \( -2x - 2z = 5 \). This is our new equation \( (5) \).

Step 4: Solve for x in Equation (5)

Rearrange equation \( (5) \) to solve for \( x \): \(-2x - 2z = 5\) becomes \(-2x = 5 + 2z\), so \(x = -\frac{5}{2} - z\).

Step 5: Substitute x from Step 4 into Equation (3)

Substitute \( x = -\frac{5}{2} - z \) into equation \( (3) \): \( (-\frac{5}{2} - z) + y + 2z = 0 \). Simplifying, it gives \( y + z = \frac{5}{2} \), so \( y = \frac{5}{2} - z \).

Step 6: Express the Solution Set

The solution set is dependent on \( z \). Use the expressions found previously: \( x = -\frac{5}{2} - z \), \( y = \frac{5}{2} - z \). The solution set in terms of \( z \) is: \[ (x, y, z) = \left(-\frac{5}{2} - z, \frac{5}{2} - z, z\right) \]

Key Concepts

Analytical Methods

When solving systems of equations, one powerful approach is using analytical methods. These methods involve finding the exact solutions through algebraic manipulations. Unlike numerical methods, which approximate solutions, analytical methods yield precise outcomes.

To solve the provided system analytically, we use a series of algebraic steps. The steps include labeling the equations for clarity, eliminating variables by combining equations, and solving for individual variables step by step. This method helps in simplifying complex systems and finding a straightforward path to the solution.

• Labeling Equations: Begins with naming each equation for reference, such as Equation (1), Equation (2), etc.


• Elimination Technique: Involves adding or subtracting equations to remove one of the variables.


• Simplification: Gradually reduces the system to simpler equations that can be easily solved.

Using these analytical methods, we can clearly derive solutions for each variable without the need for graphical representations.

Dependent Equations

Dependent equations are a type of system where the equations are not independent but rather express the same set of solutions. This means that one equation can be derived from the others by algebraic manipulation.

In the context of the given system, after eliminating the variable \( y \) and simplifying the equations, we found that all equations lead back to each other by some multiplication or addition. This is a hallmark of dependent equations. In such cases, rather than finding unique solutions, we express the solution set in terms of a free variable, which in this example is \( z \).

Understanding Dependency:

• If any equation in the system is a scalar multiple of another, the equations are dependent.


• Dependent systems often lead to infinite solutions or express solutions in terms of one or more free variables.

Recognizing dependent equations early can simplify how you approach solving the system and deciding the form of the solution set.

Solution Set

A solution set is the collection of all possible solutions that satisfy a system of equations. For systems where equations are dependent, solution sets are typically expressed using one or more parameters.

In this exercise, because the equations are dependent, we use the variable \( z \) as a parameter to express the solution set. By following the steps of rearranging and substituting variables, the system reveals its solutions in terms of \( z \).

• Parameter-based Solutions: These are used when there are infinite solutions due to dependency between equations.


• Expressing in terms of \( z \): By setting \( z \) as a free variable, we find \( x = -\frac{5}{2} - z \) and \( y = \frac{5}{2} - z \). The full solution set is thus \((x, y, z) = \left(-\frac{5}{2} - z, \frac{5}{2} - z, z\right)\).

This comprehensive view of the solution set gives us a clear understanding of how all solutions relate through the free parameter \( z \). By describing the solution set analytically, we have a complete picture of all possible solutions of the system.