Problem 42 - Algebra and Trigonometry

Question

Exercises \(41-43\) will help you prepare for the material covered in the first section of the next chapter. Solve the system: $$ \left\\{\begin{aligned} w-x+2 y-2 z &=-1 \\ x-\frac{1}{3} y+z &=\frac{8}{3} \\ y-z &=1 \\ z &=3 \end{aligned}\right. $$ Express the solution set in the form {(w, x, y, z)\\} . What makes it fairly easy to find the solution?

Step-by-Step Solution

Answer

The solution set is \{(w, x, y, z)\}. The solution is fairly easy to find because one of the unknowns, \(z\), is already given.

Step 1: Setting up the given system

The system of linear equations given is: \n \[ \begin{align*} w-x+2y-2z&=-1,\ x-\frac{1}{3} y+z&=\frac{8}{3},\ y-z&=1,\ z&=3. \end{align*} \]

Step 2: Solving for the known variable

The solution for \(z\) is given as \(z = 3\). This will be substituted into the other equations.

Step 3: Substituting the known variable

Begin by substituting \(z = 3\) in the first three equations, simplifying after each substitution. From this, we can get values for \(w\), \(x\) and \(y\).

Step 4: Obtaining the solution set

After substituting and simplifying, if we get values for all other variables, those values with the known \(z\) value forms our solution set in the form \{(w, x, y, z)\}.

Key Concepts

Substitution Method

The substitution method is an algebraic technique used to solve systems of linear equations. It involves expressing one variable in terms of others and then substituting this expression into the remaining equations. This process reduces the number of equations and variables systematically until you can find the values of all variables.

In our exercise, we immediately know the value of the variable \(z\) which is \(3\). This value is then substituted into the other three equations to find \(y\), then \(x\), and finally \(w\). This method is particularly effective when at least one equation in the system is simple enough to solve for a variable directly, as is the case with \(z=3\).

The key to using the substitution method effectively is to simplify equations as you go, reducing computational errors and making it easier to solve for the next variable. Once a variable is found, it is substituted back into the previous equations until all variable values are known.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. Solving systems of linear equations, as we've seen in the exercise, is one of the fundamental applications of linear algebra.

In our case, we have a system of four equations with four variables \((w, x, y, z)\). Linear algebra provides several methods to solve such systems including graphing, substitution, elimination, and matrix approaches like row reduction. The selected method often depends on the complexity and specific characteristics of the system.

Understanding linear algebra is crucial for many fields like engineering, physics, computer science, and more, because it provides the tools to model and solve real-world problems where relationships between variables are linear.

Ordered Quadruples

An ordered quadruple is a collection of four elements arranged in a specific order. In the context of linear algebra and specifically in systems of equations, an ordered quadruple represents the solution set with each element corresponding to one of the variables.

For example, the solution to our system of equations will be an ordered quadruple of the form \((w, x, y, z)\) where each variable is represented by a specific number denoting its value. This notation is particularly important in higher dimensions where visual representation becomes difficult or impossible.

Ordered quadruples, like ordered pairs and triples, are fundamental to the concept of vectors in linear algebra and are used widely to describe multidimensional spaces.

Variable Elimination

Variable elimination is another strategy used to solve systems of linear equations. This method involves manipulating the equations to cancel out one or more variables, reducing the system to fewer equations that are easier to solve.

This is usually achieved by adding or subtracting equations from each other after appropriately scaling them. In our exercise, the variable \(z\) is immediately eliminated because its value is given. In other systems, you might multiply an equation by a number to match the coefficient of a variable in another equation, so you can add or subtract them to eliminate that variable.

The elimination method is particularly useful when the system is too complex for simple substitution or when the coefficients of the variables align neatly, making it straightforward to eliminate variables and solve the system.