Problem 69 - A Graphical Approach to Precalculus with Limits

Question

Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} x-2 y+z &=5 \\ -2 x+4 y-2 z &=2 \\ 2 x+y-z &=2 \end{aligned}$$

Step-by-Step Solution

Answer

There is no solution; the system is inconsistent.

Step 1: Label the Equations

Label each equation in the system for reference: Equation 1: \( x - 2y + z = 5 \)Equation 2: \( -2x + 4y - 2z = 2 \)Equation 3: \( 2x + y - z = 2 \).

Step 2: Simplify Equation 2

Notice that Equation 2 can be divided by 2 to simplify it:\(-2x + 4y - 2z = 2\) becomes \(-x + 2y - z = 1\).

Step 3: Express x in terms of y and z

From the simplified Equation 2, express \(x\) in terms of \(y\) and \(z\):\(-x + 2y - z = 1\) implies \(x = 2y - z - 1\).

Step 4: Substitute x in Equations 1 and 3

Substitute \(x = 2y - z - 1\) into Equation 1: \((2y - z - 1) - 2y + z = 5\).Simplifying gives \(-1 = 5\), which isn't possible, indicating no solution exists.

Step 5: Verification

Substitute \(x = 2y - z - 1\) into Equation 3 to double-check:\(2(2y - z - 1) + y - z = 2\).This simplifies to \(4y - 2z - 2 + y - z = 2\), or \(5y - 3z = 4\), which doesn't agree with any prior results, confirming the inconsistency.

Key Concepts

Inconsistent System

An inconsistent system of equations is one in which there are no solutions. This means that the equations represent parallel lines that do not intersect at any point.
In our example, we encountered contradictions when trying to solve the system:

• Upon substituting terms, we reached an equation that stated \(-1 = 5\), which is not true.
• This indicates the system has no common solution.

In real-life scenarios, this might represent a situation where constraints simply cannot be met simultaneously. Recognizing an inconsistent system early can help avoid wasted effort and allows us to rethink assumptions or gather more information.

Linear Equations

Linear equations are algebraic expressions that represent straight lines when graphed. They typically involve constants and variables raised only to the first power.
The standard form of a linear equation is given by:

• \(ax + by + cz = d\)

In the given exercise, we worked with three linear equations that include three variables: \(x\), \(y\), and \(z\). When graphed in three dimensions, these would be planes. A solution to the system would be a point where all planes intersect.
However, if no such point exists—like in this exercise—the system is inconsistent. It's important to grasp that each linear equation contributes a plane, and their interactions define the nature of the solution: whether it's a single point, a line, or no intersection at all.

Solution Verification

Verifying solutions is a crucial part of solving systems of equations. It helps confirm whether the solutions set we obtained is correct, consistent across all original equations.
In this problem, we attempted a substitution method to verify consistency.

• From one equation, we expressed \(x\) in terms of \(y\) and \(z\).
• Upon substitution, we discovered an impossible equation \(-1 = 5\).

This verification was vital to check for errors. It also made us realize that our system of equations was inconsistent.
Inconsistent results during verification point us towards the conclusion of no solutions, helping double-check the integrity of the problem-solving process.