Problem 15 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{array}{r} x+3 y+z=6 \\ 3 x+y-z=6 \\ x-y-z=0 \end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = 3 \), \( y = 0 \), \( z = 3 \).

Step 1: Write down the system of equations

We are given the system of linear equations:\[\begin{align*}x + 3y + z &= 6 \3x + y - z &= 6 \x - y - z &= 0\end{align*}\]

Step 2: Eliminate variable from one pair of equations

First, subtract the third equation from the first equation to eliminate \( x \):\[(x + 3y + z) - (x - y - z) = 6 - 0\]This simplifies to:\[4y + 2z = 6\]Simplify by dividing the entire equation by 2:\[2y + z = 3\]

Step 3: Eliminate variable from another pair of equations

Next, eliminate \( x \) from the first and second equations.Subtract the second equation from three times the first:\[3(x + 3y + z) - (3x + y - z) = 3(6) - 6\]This becomes:\[3x + 9y + 3z - 3x - y + z = 18 - 6\]Which simplifies to:\[8y + 4z = 12\]Dividing everything by 4:\[2y + z = 3\]

Step 4: Consistency check

From Steps 2 and 3, we derived \(2y + z = 3\) twice, indicating consistency between equations, and no errors present.

Step 5: Solve for remaining variables

We now solve for \( y \) and \( z \) using \(2y + z = 3\).Choose \( y = 0 \):\[2(0) + z = 3\]\[z = 3\]Substitute \( y = 0 \) and \( z = 3 \) into \( x = y + z \):\[x = 0 + 3 = 3\]

Step 6: Verify the solution

Substitute \( x = 3 \), \( y = 0 \), and \( z = 3 \) into the original equations to verify.1. \(3 + 3(0) + 3 = 6\) ✓2. \(3(3) + 0 - 3 = 6\) ✓3. \(3 - 0 - 3 = 0\) ✓All equations are satisfied, thus, the solution is correct.

Key Concepts

Equation Solving

Solving a system of equations involves finding the set of values for the variables that satisfy all the given equations simultaneously. For the linear system provided, the main goal is to use algebraic manipulations to isolate the variables, typically starting with one at a time. Each equation is like a condition that the variables need to meet. When combined, all these conditions define a unique solution for the system or reveal that no solution exists. Solving starts with writing down all equations and carefully transforming them without breaking the equality.
For this system, the equations are given as:

• \( x + 3y + z = 6 \)
• \( 3x + y - z = 6 \)
• \( x - y - z = 0 \)

Each equation contains the variables \(x\), \(y\), \(z\). Solving the system means determining values for these variables that satisfy all three conditions.

Variable Elimination

Variable elimination is a powerful technique to simplify a system of equations by removing one of the variables. The idea is to manipulate the equations such that when two or more of them are combined, one variable disappears. This reduces the complexity of the system.
The initial step often includes selecting pairs of equations to eliminate a chosen variable. In the given exercise, the variable \( x \) was targeted. By subtracting and transforming equations, two simplified equations with only \( y \) and \( z \) remained.

• Subtracting the third equation from the first gave \( 4y + 2z = 6 \)
• Subtracting the second from three times the first produced the equation \( 8y + 4z = 12 \), which is simplified to \( 2y + z = 3 \)

This process reduced the system to a simpler form, enabling easier solving.

Consistency Check

A consistency check is crucial to ensure that simplified equations correctly represent the original system. It essentially confirms whether the manipulations logically led to a valid solution or consistent result. In the exercise, after eliminating variables, both derived equations were the same: \( 2y + z = 3 \).
Such a finding indicates that the equations are consistent and interconnected, reaffirming that no contradictions were introduced. With consistency confirmed, solving for the remaining variables is straightforward.

• Choosing any simple value for one of the variables (like \( y = 0 \)) allows solving the other (\( z = 3 \)).
• Substituting back gives \( x = 3 \).

The consistency check ensures that these values satisfy all original equations, confirming the solution's correctness.