Problem 16 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{array}{lr} 2 x-y+2 z= 6 \\ -x+y+z= 0 \\ -x \quad \quad -3 z=-6 \end{array} $$

Step-by-Step Solution

Answer

The system is consistent with a general solution: \((x, y, z) = (-3z+6, -4z+6, z)\).

Step 1: Write the System of Equations

We start with the given system of equations:1. \(2x - y + 2z = 6\)2. \(-x + y + z = 0\)3. \(-x - 3z = -6\).

Step 2: Solve for x from Equation 3

Solve for \(x\) in the third equation:\[-x - 3z = -6 \quad \Rightarrow \quad x = -3z + 6\].

Step 3: Substitute x in Equations 1 and 2

Substitute \(x = -3z + 6\) into the first and second equations.For Equation 1:\[2(-3z + 6) - y + 2z = 6\]\[-6z + 12 - y + 2z = 6\]\[-4z - y = -6\quad \Rightarrow \quad y = -4z + 6\].For Equation 2:\[-(-3z + 6) + y + z = 0\]\[3z - 6 + y + z = 0\]\[y + 4z = 6\quad \Rightarrow \quad y = -4z + 6\].

Step 4: Consistency Check

Both substitutions for \(y\) give the same expression, \(y = -4z + 6\), confirming consistency. Now, substitute back to find a particular solution. For simplicity, let's assume an arbitrary value for \(z\).

Step 5: Find Particular Solution

Let \(z = 0\). Then:\[x = -3(0) + 6 = 6\]\[y = -4(0) + 6 = 6\].Thus, one solution is \((x, y, z) = (6, 6, 0)\).

Step 6: General Solution

Since the arbitrary choice of \(z\) gave us consistency in steps, the solutions are given by:\[x = -3z + 6, \quad y = -4z + 6, \quad z = z\].For any real number \(z\), these expressions will satisfy the system.

Key Concepts

System of Equations

When discussing linear algebra, a system of equations is like having a set of puzzles where each variable represents a piece of the solution. A system of equations consists of multiple equations that are all connected by their variables, and our task is to find the values for these variables that satisfy all equations simultaneously. In the example from the original exercise, you're presented with three linear equations containing the variables, \(x\), \(y\), and \(z\):

• \(2x - y + 2z = 6\)
• \(-x + y + z = 0\)
• \(-x - 3z = -6\)

Each equation represents a condition that the variables must meet. To "solve" the system of equations means to find a set of values for \(x\), \(y\), and \(z\) that make all three conditions true at once. A solution can be a single point, show a pattern for infinite solutions, or indicate no solution exists if the equations are inconsistent.

Linear Equations

Linear equations are the building blocks of a system of equations in linear algebra. A linear equation is one where variables appear only to the first power and are not multiplied together. For instance, in the exercise, equations like \(2x - y + 2z = 6\) and \(-x - 3z = -6\) are examples of linear equations. These equations can represent geometric lines or planes when plotted graphically.

• In 2D, a single linear equation appears as a straight line.
• In 3D, as in our exercise, each linear equation appears as a plane.

The goal is often to find the intersection point(s) of these lines or planes, as the intersection represents the solution where all conditions are met. Understanding linear equations helps you visualize what's happening geometrically and shows how alterations in coefficients and constants affect the position and orientation of these lines or planes.

Solution Techniques

Solving a system of linear equations involves several techniques that help find the values of the variables that satisfy all equations in the system. The most common methods include:

• Substitution Method: This involves solving one of the equations for a specific variable and substituting that expression into the other equations. This method simplifies the system step by step. As in the exercise provided, solving for \(x\) in the third equation as \(x = -3z + 6\), and then substituting into the other equations reduces the complexity.
• Elimination Method: By adding or subtracting equations, this method eliminates one variable at a time. It works well when you have equations that can be easily combined or transformed to cancel out variables.
• Matrix Methods: Finally, for larger systems, matrix operations such as Gaussian elimination are very effective. These involve representing the system in matrix form and using operations to reduce the matrix to easily identifiable solutions.

Each technique provides strategies to move from a potentially overwhelming number of equations down to a more manageable problem, ultimately leading to one or more solutions. In systems with more variables than equations, solutions can often be expressed in terms of free variables, showing infinite solutions, as demonstrated by the general solution provided in the exercise.