Problem 14 - College Algebra with Modeling and Visualization

Question

If possible, solve the system. $$ \begin{array}{rr} -x-5 y+2 z= & 2 \\ x+y+2 z= & 2 \\ 3 x+y-4 z= & -10 \end{array} $$

Step-by-Step Solution

Answer

\(x = -\frac{3}{2}, y = \frac{1}{2}, z = \frac{3}{2}\).

Step 1: Write the System of Equations

First, write the system of equations clearly: \(-x - 5y + 2z = 2\), \(x + y + 2z = 2\), \(3x + y - 4z = -10\). Our goal is to find the values of \(x\), \(y\), and \(z\).

Step 2: Add Equations to Eliminate x

Add the first equation to the second equation. This will eliminate \(x\): \((-x - 5y + 2z) + (x + y + 2z) = 2 + 2\). The result is \(-4y + 4z = 4\) or \(-y + z = 1\) (dividing everything by 4).

Step 3: Add Equations to Eliminate x Again

Add the first equation to the third equation to eliminate \(x\): \((-x - 5y + 2z) + (3x + y - 4z) = 2 - 10\). Simplify to get \(2x - 4y - 2z = -8\), which simplifies to \(x - 2y - z = -4\) (dividing everything by 2).

Step 4: Substitute to Eliminate z

Now you have two equations: \(-y + z = 1\) and \(x - 2y - z = -4\). Substitute \(z = y + 1\) into the second equation. This gives \(x - 2y - (y + 1) = -4\) which simplifies to \(x - 3y - 1 = -4\) and further to \(x = 3y - 3\).

Step 5: Simplify the System

Now, substitute \(x = 3y - 3\) and \(z = y + 1\) back into one of the original equations, such as \(x + y + 2z = 2\): Substitute to get:\((3y - 3) + y + 2(y + 1) = 2\).Simplify to find \(3y - 3 + y + 2y + 2 = 2\), Leading to \(6y - 1 = 2\), which simplifies to \(6y = 3\), and further to \(y = \frac{1}{2}\).

Step 6: Find x and z

Substitute \(y = \frac{1}{2}\) into \(x = 3y - 3\) to find \(x = 3(\frac{1}{2}) - 3 = \frac{3}{2} - 3 = -\frac{3}{2}\). Next, use \(z = y + 1\) to get \(z = \frac{1}{2} + 1 = \frac{3}{2}\).

Step 7: Verify the Solution

Substitute \(x = -\frac{3}{2}\), \(y = \frac{1}{2}\), and \(z = \frac{3}{2}\) back into the original equations to verify the solution. Check each equation to ensure both sides are equal.

Key Concepts

Solving Linear Systems

A linear system of equations consists of two or more linear equations that share common variables. Solving these systems involves finding the values of the variables that make all the given equations true simultaneously.

Typically, the components of a linear equation are written in the standard form:

• The coefficients of variables.
• A constant term.

To solve a system of equations means finding values for each variable that satisfy each equation in the system. The methods to solve these systems include graphing, substitution, elimination, and using matrices. Graphical solutions can provide a visual representation, but algebraic methods are more precise, especially for complex systems.

Substitution Method

The substitution method is one of the primary algebraic techniques to solve systems of equations. In this method:

• You solve one of the equations for one variable in terms of others.
• Substitute this expression into the other equation(s). This reduces the number of equations, allowing you to solve for the remaining variables.

For example, consider substituting in a system where you obtain a unique solution for a variable:
1. Solve \( z = y + 1 \) and substitute it into another equation like \(x - 2y - z = -4\), to reduce the system to simpler terms. 2. This manipulates the system step-by-step until you identify explicit solutions for each variable.

By continuously substituting and simplifying, you steadily reduce the complexity of the system. The substitution method is particularly effective when one equation is easy to solve for one of the variables.

Elimination Method

The elimination method, sometimes called the "addition method," involves combining equations to eliminate one variable, making it easier to solve the system. This method usually involves:

• Adding or subtracting equations to cancel out one variable.
• Simplifying the resulting equation to solve for another variable.

In our exercise, elimination is demonstrated by combining the first two equations to eliminate the \(x\) variable:
- By adding the equations \( (-x - 5y + 2z) + (x + y + 2z) = 4 \), you effectively remove \(x\), simplifying to \( -y + z = 1 \).
This method is highly efficient for solving systems where coefficients can easily be manipulated for cancellation. It provides a straightforward approach to handling more complex systems by systematically reducing them.

Verification of Solutions

Verifying the solutions of a system of equations involves substituting the found values back into the original equations to ensure they satisfy all equations in the system. This step is crucial to confirm the correctness of your solutions.

For example, if you solve a system and find \(x = -\frac{3}{2} \), \(y = \frac{1}{2} \), and \(z = \frac{3}{2} \), verification involves:

• Plugging these values back into each original equation.
• Checking that each equation holds true (i.e., both sides of each equation are equal).

If any equation does not verify, it indicates a mistake in calculations, requiring a review of the solutions. Verification ensures that all solutions are accurate and consistent with the given system.