Problem 15 - Precalculus: Mathematical for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} x+y+z &=4 \\ x+3 y+3 z &=10 \\ 2 x+y-z &=3 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system is consistent with solution: \( x = 2z-1, y = 3-z, z = z \).

Step 1: Write the System of Equations

The given system of linear equations can be written as: 1. \( x + y + z = 4 \) 2. \( x + 3y + 3z = 10 \) 3. \( 2x + y - z = 3 \).

Step 2: Eliminate x from Equations 2 and 3

Subtract Equation 1 from Equation 2 to eliminate \( x \). \[(x + 3y + 3z) - (x + y + z) = 10 - 4 \2y + 2z = 6 \Divide the entire equation by 2:\y + z = 3 \]

Step 3: Eliminate x from Equations 1 and 3

Subtract Equation 1 from Equation 3 to also eliminate \( x \).\[(2x + y - z) - (x + y + z) = 3 - 4 \x - 2z = -1 \]

Step 4: Solve the Two-Variable System

We now solve the following system derived from previous steps: 1. \( y + z = 3 \)2. \( x - 2z = -1 \)From equation 1: \( y = 3 - z \).Substitute \( y = 3 - z \) in equation 2:\[x - 2z = -1\]

Step 5: Express Variables in Terms of z

Substituting \( y = 3 - z \) in original Equation 3 yields expressions for \( x \):Substitute \( y = 3-z \) in:\\[x - 2z = -1 \x = 2z - 1 \]

Step 6: General Solution

The general solution for the system is found as:\[x = 2z - 1 \y = 3 - z \z = z \]This system is consistent and the general parametric solution is obtained by letting \( z = t \), where \( t \) is a parameter.

Key Concepts

Inconsistent Systems

When dealing with systems of linear equations, it's crucial to determine whether the system is consistent or inconsistent. An inconsistent system is one that has no solution. This happens because the equations contradict each other.
For our system, after applying elimination techniques to simplify the equations, we can check for contradictions. If at any point an equation simplifies to an untrue statement, such as \(0 = 5\), the system is inconsistent.
In the exercise example, the system ended with consistent equations, which means there is at least one solution or potentially infinitely many.

Parametric Solutions

Parametric solutions arise in systems of equations that have more variables than independent equations. This results in infinitely many solutions. Each solution is represented by a particular choice of the parameters.
In our solved exercise, the general solution was expressed with \(z = t\), allowing infinite options for \(t\). By varying \(t\), we create solutions for \(x\) and \(y\) as well:

• \(x = 2t - 1\)
• \(y = 3 - t\)
• \(z = t\)

This shows that different values of \(t\) lead to different sets of solutions, expressing dependence on one chosen parameter.

Variable Elimination

Variable elimination is a technique to simplify systems of equations by removing variables. It helps to focus on fewer variables to solve the system more easily.
In our exercise, we used elimination to first remove \(x\) from the equations, reducing the 3-variable system to a 2-variable system.
By subtracting equations strategically, we arrived at simpler equations involving just two variables \(y\) and \(z\), or \(x\) and \(z\), which can then be solved to express variable dependencies.
Eliminating and reducing variables aims to eventually express all variables in terms of a parameter, enabling us to find a parametric solution for systems with infinite solutions.