Problem 23 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{array}{rr} 3 x+2 y+z= & -1 \\ 3 x+4 y-z= & 1 \\ x+2 y+z= & 0 \end{array} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = -\frac{1}{2}\), \(y = \frac{1}{2}\), \(z = -\frac{1}{2}\).

Step 1: Write Down the System of Equations

We are given a system of three equations:\[ \begin{align*} 3x + 2y + z &= -1, \3x + 4y - z &= 1, \x + 2y + z &= 0.\end{align*} \] Our goal is to find values of \(x\), \(y\), and \(z\) that satisfy all three equations.

Step 2: Subtract Equations to Eliminate a Variable

Subtract the third equation from the first equation to eliminate \(x\):\[ (3x + 2y + z) - (x + 2y + z) = -1 - 0 \]This simplifies to:\[ 2x = -1 \]Thus, we find:\[ x = -\frac{1}{2} \]

Step 3: Use Back-Substitution to Find Another Variable

Substitute \(x = -\frac{1}{2}\) into the third equation:\[ (-\frac{1}{2}) + 2y + z = 0 \]Solve for \(z\):\[ 2y + z = \frac{1}{2} \]Rearrange to find \(z\):\[ z = \frac{1}{2} - 2y \]

Step 4: Substitute to Solve for Remaining Variables

Substitute \(x = -\frac{1}{2}\) into the second equation:\[ 3(-\frac{1}{2}) + 4y - z = 1 \]Simplify and solve for \(y\):\[ -\frac{3}{2} + 4y - z = 1 \]\[ 4y - z = 1 + \frac{3}{2} \]\[ 4y - z = \frac{5}{2} \]Now substitute \(z = \frac{1}{2} - 2y\) into this equation:\[ 4y - (\frac{1}{2} - 2y) = \frac{5}{2} \]\[ 4y - \frac{1}{2} + 2y = \frac{5}{2} \]\[ 6y - \frac{1}{2} = \frac{5}{2} \]Solve for \(y\):\[ 6y = \frac{5}{2} + \frac{1}{2} = 3 \]\[ y = \frac{3}{6} = \frac{1}{2} \]

Step 5: Determine the Final Variable

Using \(y = \frac{1}{2}\), substitute back to find \(z\):\[ z = \frac{1}{2} - 2(\frac{1}{2}) \]\[ z = \frac{1}{2} - 1 = -\frac{1}{2} \]

Step 6: Verify the Solution

Substitute \(x = -\frac{1}{2}\), \(y = \frac{1}{2}\), and \(z = -\frac{1}{2}\) back into all three of the original equations to ensure they are satisfied:1. \(3(-\frac{1}{2}) + 2(\frac{1}{2}) + (-\frac{1}{2}) = -1\) ✔2. \(3(-\frac{1}{2}) + 4(\frac{1}{2}) - (-\frac{1}{2}) = 1\) ✔3. \((-\frac{1}{2}) + 2(\frac{1}{2}) + (-\frac{1}{2}) = 0\) ✔Since all three equations are satisfied, the solution is correct.

Key Concepts

Linear Algebra

Linear Algebra is a branch of mathematics focused on vectors, vector spaces (also known as linear spaces), linear transformations, and systems of linear equations. In linear algebra, we often deal with equations like the system provided in the exercise. These involve several unknowns, and the goal is to find the values that satisfy all the given equations simultaneously.

The systems of equations can be solved using different techniques, such as graphing, substitution, and elimination. Linear algebra is fundamental in various scientific domains, influencing fields like computer science, engineering, physics, and economics. It provides techniques for solving real-world problems using mathematical models.

Elimination Method

The elimination method is an efficient way to solve systems of equations by eliminating one variable to solve for the others. This method involves adding or subtracting the equations.

The goal is to simplify the system, crafting new equations that suppress one variable, making it easier to find others.

• First, identify a variable to be eliminated (often chosen for simplicity).
• Perform operations (addition or subtraction) on pairs of equations to derive a new system of equations where one variable is missing.

In the exercise, by subtracting the third equation from the first, we eliminated the variable y, making it significantly easier to solve for x. Elimination is a highly effective method because it reduces the complexity of the system step by step.

Back-Substitution

Back-substitution is a method employed after applying either substitution or elimination to solve systems of equations.

Once a simpler form of the system is achieved, typically after elimination, back-substitution involves solving for one variable and substituting that back into other equations to unravel the remaining variables.

• Begin with the last equation in the series that contains only one variable.
• Use this solution to substitute back into previous equations, gradually solving for additional variables.

In our stepwise solution, after eliminating a variable using the elimination method, we used back-substitution to successively solve for z and then y, ensuring every variable's value was correctly deduced.

Solution Verification

Verifying the solution to a system of equations is crucial to ensure the accuracy of results.

Verification involves substituting the found values back into the original equations to confirm they hold true. This not only checks for correctness but can also uncover any errors during the solving stages.

• Substitute each variable's value back into the original equations step-by-step.
• Ensure that each equation balances correctly, meaning both sides are equal.

In the exercise's final step, we verified our solution by plugging the values of x, y, and z back into each of the three original equations, confirming that all were satisfied, ensuring our solution was indeed correct.