Problem 14 - Algebra and Trigonometry

Question

\(13-16\) . Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. $$ \begin{array}{l}{\left\\{\begin{aligned} x+y-3 z &=3 \\\\-2 x+3 y+z &=2 \\\ x-y+2 z &=0 \end{aligned}\right.} \\ {\text { Eliminate the } x \text { -term }} \\ {\text { from the second equation. }}\end{array} $$

Step-by-Step Solution

Answer

The new system is \( \{ x+y-3z=3, \; 5y-5z=8, \; x-y+2z=0 \} \).

Step 1: Identify the Equation to Modify

We want to eliminate the \( x \) term in the second equation, namely \(-2x + 3y + z = 2\). Our goal is to use another equation to achieve this.

Step 2: Choose an Equation for Elimination

We will use the first equation, \( x + y - 3z = 3 \), to eliminate the \( x \) term in the second equation. This is because the first equation has a \( +x \) term which will help us eliminate \( x \) when combined with the \( -2x \) term.

Step 3: Scale the Chosen Equation

Multiply the first equation by 2 to align the coefficient of \( x \) for elimination:\[ 2(x + y - 3z) = 2(3) \]This simplifies to:\[ 2x + 2y - 6z = 6 \]

Step 4: Add the Modified Equation to the Target Equation

Add this scaled version of the first equation to the second equation to eliminate the \( x \) term:\[ (2x + 2y - 6z) + (-2x + 3y + z) = 6 + 2 \]The resulting equation is:\[ 5y - 5z = 8 \]

Step 5: Write the New System

Substitute the new equation \( 5y - 5z = 8 \) back into the system, which becomes:\[\begin{align*}x + y - 3z &= 3 \5y - 5z &= 8 \x - y + 2z &= 0\end{align*}\]

Key Concepts

Elimination Method

The elimination method is a strategy used to solve systems of linear equations. The main idea is to add or subtract equations in such a way that one variable is eliminated, allowing you to solve for the remaining variables. This method often involves scaling one or more equations so that when they are added or subtracted, one variable cancels itself out. This approach is particularly helpful when dealing with systems of equations where direct substitution is cumbersome or results in complex expressions.

Here's how the elimination method works in simple steps:

• Choose one of the variables to eliminate. Look for opportunities where the coefficients of a variable are opposites or can easily be made into opposites by multiplying the entire equation by a number.
• Scale one or both equations, if necessary, so that the coefficients of the chosen variable are equal and opposite. This often involves basic arithmetic operations like multiplication or division.
• Add or subtract the equations to eliminate the chosen variable. This will result in a new equation with one fewer variable.
• Solve the simplified system, which now has one less equation and one less variable.
• Back-substitute the found value(s) into one of the original equations to solve for the remaining variables.

This method is systematic and can be applied to any linear system, making it a versatile tool in linear algebra.

Systems of Equations

Systems of equations involve a set of two or more equations with the same variables. The objective is to find the values for these variables that satisfy all of the equations simultaneously. This concept is foundational in algebra and is used to model real-world situations where multiple conditions must be satisfied at once.

There are several methods to solve systems of equations:

• Graphical Method: Solving the system by plotting the equations on a graph and finding the intersection points. This may not always be precise, especially for complex systems.
• Substitution Method: Solving one of the equations for one variable and substituting the result into the other equation. This method is straightforward but can become cumbersome with more complex systems.
• Elimination Method: As explained earlier, this involves adding or subtracting equations to eliminate a variable. It is often more systematic and can be easily executed without visually plotting or complex algebra manipulations.

In a system of linear equations, the relationships between the variables are expressed as linear equations, each corresponding to a line in a plane. The solution to the system is typically the point where these lines intersect, representing the values that satisfy all equations simultaneously.

Linear Algebra

Linear algebra is a branch of mathematics concerned with vectors, spaces, and linear mappings between such spaces. One of its central concepts is solving systems of linear equations. These are crucial as they describe linear relationships, which are fundamental in various scientific and engineering disciplines.

Key concepts in linear algebra related to solving systems of equations include:

• Vector Spaces: The set of all possible solutions to a system of linear equations forms a vector space. Understanding this helps in visualizing how solutions relate to each other.
• Matrices and Determinants: These are used to represent and solve linear systems efficiently. Matrices can be manipulated to simplify systems, and their determinants can indicate the existence and uniqueness of solutions.
• Matrix Inversion and Row Reduction: These techniques are used to solve systems more systematically, particularly in computational applications where numerous variables are involved.

Linear algebra provides the tools to handle large systems of equations that appear in various applications, from computer science to physics. It abstracts problems to a form that can be solved analytically and computationally, offering insights into the mechanics of the underlying systems.