Problem 64 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{aligned} y-z+2 w &=0 \\ 3 x+2 y &+w=0 \\ 2 x &+4 w=12 \\ -2 x &-2 z+5 w=6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution depends on \(w\), with expressions: \(x = 6 - 2w; y = \frac{5w - 18}{2}; z = \frac{9w - 18}{2}\).

Step 1: Identify the equations

We have a system of four linear equations involving four variables: \(x, y, z,\) and \(w\). The equations are: 1. \( y - z + 2w = 0 \)2. \( 3x + 2y + w = 0 \)3. \( 2x + 4w = 12 \)4. \( -2x - 2z + 5w = 6 \)

Step 2: Solve for a variable

Start finding one of the variables from one of the simpler equations. From Equation (3), \( 2x + 4w = 12 \), solve for \(x\):\( x = 6 - 2w \).

Step 3: Substitute x in another equation

Substitute \( x = 6 - 2w \) in Equation (2), \( 3x + 2y + w = 0 \):\( 3(6 - 2w) + 2y + w = 0 \)\( 18 - 6w + 2y + w = 0 \)\( 2y - 5w = -18 \)\( y = \frac{5w - 18}{2} \).

Step 4: Substitute x and y into z equation

Substitute \( x = 6 - 2w \) and \( y = \frac{5w - 18}{2} \) into Equation (1), \( y - z + 2w = 0 \):\( \frac{5w - 18}{2} - z + 2w = 0 \)\( 5w - 18 + 4w = 2z \)\( 9w - 18 = 2z \)\( z = \frac{9w - 18}{2} \).

Step 5: Substitute x, y, z into the last equation

Substitute \( x = 6 - 2w \), \( y = \frac{5w - 18}{2} \), and \( z = \frac{9w - 18}{2} \) into Equation (4), \(-2x - 2z + 5w = 6 \):\( -2(6 - 2w) - 2\left(\frac{9w - 18}{2}\right) + 5w = 6 \)\( -12 + 4w - (9w - 18) + 5w = 6 \)\( -12 + 4w - 9w + 18 + 5w = 6 \)\( -12 + 18 + 0w = 6 \)\( 6 = 6 \), which is true for any \(w\).

Step 6: Determine the solution set

Since \(w\) is a free variable, choose any value for \(w\) to find the corresponding \(x, y,\) and \(z\). For example, if \(w = 0\):- \( x = 6 - 2(0) = 6 \)- \( y = \frac{5(0) - 18}{2} = -9 \)- \( z = \frac{9(0) - 18}{2} = -9 \).

Step 7: General solution

The system is dependent on \(w\), and the general solutions are: - \( x = 6 - 2w \)- \( y = \frac{5w - 18}{2} \)- \( z = \frac{9w - 18}{2} \) - \( w = w \)where \(w\) can be any real number.

Key Concepts

Understanding Variables in a Linear System

Variables are essential elements in a mathematical equation, representing unknowns that we aim to solve. In the context of the given system of linear equations, our variables are \(x, y, z,\) and \(w\). These symbols stand in for quantities that satisfy each equation simultaneously.
When dealing with a linear system, we often have more equations, or equally, just as many, as the number of variables. Our goal is to find a set of values for these variables that make all the equations true at the same time. Approaching this requires using algebraic methods such as substitution, like we did in the example solution. Here, we expressed \(x\) as \(x = 6 - 2w\) to minimize complexity in other equations and steadily solve for the rest.
Understanding the role of variables and their interdependence is essential in navigating through any linear system of equations.

Exploring the Solution Set

A solution set includes all possible answers that satisfy each equation in the system. In our problem, after resolving the relationship between all variables, we discover that \(w\) is a free variable.
Simply put, a free variable means that it can take any value, leading to multiple solutions. The solution set then becomes expressed in terms of \(w\), allowing for infinite values as long as they satisfy the given linear equations.

• For example, with \(w = 0\), we find specific values: \(x = 6\), \(y = -9\), \(z = -9\).
• Choose a different value of \(w\), and you'll find another valid solution.

Understanding the solution set allows one to see the breadth of possibilities within a linear system and can define infinite solutions that span a one-dimensional space when visualized graphically.

Dependent System Explained

A dependent system arises when the equations in a linear system do not uniquely identify one solution. Instead, there are infinite solutions, often depending on one or more "free" variables. In the case of our exercise, the system depends on the variable \(w\).
When solving, you find that \(x, y, z\) depend entirely on \(w\). This means one variable is free and drives the value of the others upstream, creating a dependent system.
The presence of a dependent system suggests some equations are combinations of others, reducing the effective dimension or instruction the system provides. This is reflected in the consistent or overlapping nature when seen as graphs, like lines or planes occupying the same space.

How Systems of Equations Work

A system of equations consists of multiple mathematical statements that work together. Each equation provides a specific "rule" that the variables must satisfy. Solving such systems involves finding all variable values that meet all conditions simultaneously.
For linear systems, these rules appear as straight-line relationships amongst variables. The process often involves methods such as:

• Substitution, where one expresses a variable in terms of others to simplify and reduce the system step-by-step.
• Elimination, where adding or subtracting equations help to eliminate variables and solve the system gradually.

Linear systems can offer unique solutions (one set of values), no solution (when equations contradict), or infinite solutions like our dependent system, where equations are interrelated such that multiples of one variable form valid solutions.