Problem 67 - A Graphical Approach to Precalculus with Limits

Question

Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} x+2 y-z &=0 \\ 3 x-y+z &=6 \\ -2 x-4 y+2 z &=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is: \(x = \frac{12 - z}{7}\), \(y = \frac{-6 + 4z}{7}\), \(z = z\).

Step 1: Analyze the System

We are given a system of three linear equations with three variables: \(x\), \(y\), and \(z\). The equations are:\[\begin{align*}x + 2y - z &= 0, \3x - y + z &= 6, \-2x - 4y + 2z &= 0\end{align*}\]We'll solve this system using substitution or elimination methods. It appears that combining equations to eliminate a variable could work efficiently.

Step 2: Simplify the Third Equation

Notice that the third equation, \[-2x - 4y + 2z = 0\]can be divided by \(-2\) to simplify:\[x + 2y - z = 0\]This is actually identical to the first equation.

Step 3: Substitute Equation Reduction

Since the third equation is a multiple of the first, we focus on solving the system of two equations: \(x + 2y - z = 0\) and \(3x - y + z = 6\). We'll eliminate \(z\) by adding these two equations together:\[(x + 2y - z) + (3x - y + z) = 0 + 6\]This simplifies to:\[4x + y = 6\]

Step 4: Solve for y

We now have an equation in terms of \(x\) and \(y\):\[4x + y = 6\]Solve for \(y\):\[y = 6 - 4x\]

Step 5: Back Substitute to Find x and z

Use \(y = 6 - 4x\) in the first equation \(x + 2y - z = 0\):\[x + 2(6 - 4x) - z = 0\]Simplifying gives:\[x + 12 - 8x - z = 0\]\[-7x + 12 - z = 0\]Solve for \(z\):\[z = 12 - 7x\]

Step 6: Express Solution in Terms of z

Since \(z\) is derived as \(z = 12 - 7x\), express other variables in terms of \(z\):\[7x = 12 - z \x = \frac{12 - z}{7}\]Substitute \(x = \frac{12 - z}{7}\) into \(y = 6 - 4x\):\[y = 6 - 4\left(\frac{12 - z}{7}\right)\]Further simplify:\[y = 6 - \frac{48 - 4z}{7}\]Combine and simplify:\[y = \frac{42 - 48 + 4z}{7} = \frac{-6 + 4z}{7}\]

Step 7: Final Solution

The system of equations has infinite solutions given in terms of \(z\):\[x = \frac{12 - z}{7}, \y = \frac{-6 + 4z}{7}, \z = z\]

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of linear equations. It involves solving one equation for one variable and then substituting this expression into the other equations. This can simplify the system and make it easier to solve.
Let's consider the system of equations provided:

• First, identify an equation that can be easily solved for one of its variables. In our case, all equations give a straightforward start.
• Then, replace the chosen variable in the other equations with the expression you've found.
• This reduces the number of equations you have and often transforms the system into a simpler form.

By substituting one equation into another, we can find specific expressions for all variables. This method is especially useful when one equation is already nearly solved for a particular variable.

Elimination Method

The elimination method, also known as the addition method, is another effective technique for solving systems of linear equations. It involves adding or subtracting equations to eliminate a variable, hence the name.
In our exercise, notice how:

• We can combine equations to eliminate one variable, which simplifies the system.
• For example, by adding the modified first and second equations, we eliminated the variable \(z\), reducing complexity.

The key here is to align the equations so that adding or subtracting them will remove one of the variables. This simplifies solving for the remaining variables. The elimination method becomes very efficient when the coefficients allow for easy removal of variables.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It’s essential for solving systems of linear equations as it provides systematic methods, like substitution and elimination, to find solutions.
Key concepts in linear algebra include:

• Systems of Equations: Collections of linear equations involving the same set of variables.
• Matrices: Rectangular arrays of numbers which can represent linear equations.
• Vectors: Objects that have both a direction and magnitude, often used for clarity in solutions.

These core ideas facilitate the manipulation of equations and can help find solutions, detect special cases like infinite solutions, and apply transformations to systems effectively.

Infinite Solutions

A system of equations may have infinitely many solutions if the equations describe the same geometric object, such as a line or plane, in space. This happens when:

• Equations are identical or scalar multiples of each other.
• You simplify equations and end up with one or more dependent equations.
• There’s not enough independent information to isolate unique points.

In our solved example, the existence of infinite solutions is determined by expressing solutions in terms of a free variable (\(z\)). When the solution is dependent on a parameter, it indicates that all variables can vary indefinitely while still satisfying all original equations.