Problem 13 - Precalculus Mathematics for Calculus

Question

Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. $$\left\\{\begin{array}{r} x-2 y-z=4 \\ x-y+3 z=0 \\ 2 x+y+z=0 \end{array}\right.$$

Step-by-Step Solution

Answer

New system without \( x \): \( \{ \begin{array}{l} y + 4z = -4 \\ 3x + 4y + 3z = -4 \end{array} \} \).

Step 1: Identify the System of Equations

The system of equations provided is as follows: \( \{ \begin{array}{l} x - 2y - z = 4 \ x - y + 3z = 0 \ 2x + y + z = 0 \end{array} \} \). We need to eliminate one of the variables, let's choose to eliminate \( x \).

Step 2: Eliminate \( x \) in the First and Second Equations

Subtract the first equation from the second equation to eliminate \( x \): \[ (x - y + 3z) - (x - 2y - z) = 0 - 4 \]Simplifying, we have:\[ (x - x) + (-y + 2y) + (3z + z) = -4 \]\[ y + 4z = -4 \]

Step 3: Eliminate \( x \) in the First and Third Equations

Subtract the first equation from the double of the third equation to also eliminate \( x \):\[ 2(2x + y + z) - (x - 2y - z) = 2(0) - 4 \]Simplifying, we obtain:\[ 4x + 2y + 2z - x + 2y + z = -4 \]\[ (4x - x) + (2y + 2y) + (2z + z) = -4 \]\[ 3x + 4y + 3z = -4 \]

Step 4: Write the New Equivalent System

The system of equations with the variable \( x \) eliminated is:\[ \{ \begin{array}{l} y + 4z = -4 \ 3x + 4y + 3z = -4 \end{array} \} \]. This system excludes the first equation now as it was used to derive the other equations, therefore \( x \) was effectively eliminated from the set.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In other words, each equation represents a line, plane, or higher-dimensional entity. Solving a system means finding all variable values that satisfy all the equations at once.

When solving systems, you may encounter different scenarios:

• Unique solution: Where the lines or planes intersect at a single point.
• Infinite solutions: Where the lines or planes overlap completely.
• No solution: Where the lines or planes never meet.

The goal in working with a system of equations is often to find an operational technique that can easily and systematically bring about the solution set if it exists. One such method is the "Elimination Method," which can help in matters of clarity when one would like to work step-by-step in solving the problem.

Linear Algebra

Linear algebra focuses on vector spaces and linear mappings between them. It is forgiving in terms of its practical usefulness in developing techniques such as those used to solve systems of linear equations.

Key components in linear algebra include:

• Vectors: Quantities with both magnitude and direction, which can be visualized as points or arrows in space.
• Matrices: Rectangular arrays of numbers that can represent systems of linear equations.
• Determinants: A special number that can be calculated from a matrix, important in matrix operations.

Using linear algebraic methods, one can manipulate matrices and vectors to simplify and resolve systems of equations effectively. For instance, elimination methods are derivative designs of linear algebra, providing mechanisms to systematically reduce variables in equations.

Variable Elimination

Variable elimination is a method employed in solving systems of equations, where specific variables are removed to simplify the system. The advantages of using elimination include its systematic nature and relative simplicity in execution.

Steps to eliminate a variable:

• Choose a variable: Pick a variable you wish to eliminate in order to simplify the system.
• Combine equations: Add or subtract equations to cancel out the chosen variable from your system.
• Simplify: Once eliminated, simplify the remaining equations to solve for other variables.

The original problem is an excellent example of this technique in action. By systematically using subtraction and combination of equations, a variable (in this case, \(x\)) was successfully removed, leading to a simpler system. This approach makes it much easier to identify the solutions to the remaining equations, as the system is thus reduced in complexity.