Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

\(\left\\{\begin{array}{r}x+y-2 z=0 \\ x-y-4 z=0 \\ y+z=0\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (3t, -t, t)\) for any real number \(t\).

Step 1: Write the System of Equations

Consider the system of linear equations given by: \[\begin{align*}x + y - 2z &= 0 \x - y - 4z &= 0 \y + z &= 0 \end{align*}\]

Step 2: Solve the Third Equation for a Variable

From the third equation \( y + z = 0 \), we can solve for \( y \) in terms of \( z \): \[ y = -z \]

Step 3: Substitute into the First Equation

Substitute \( y = -z \) from Step 2 into the first equation: \[ x + (-z) - 2z = 0 \]Simplify this to get: \[ x - 3z = 0 \]Thus, \( x = 3z \).

Step 4: Substitute into the Second Equation

Using \( y = -z \) and \( x = 3z \) from previous steps, substitute into the second equation: \[ 3z - (-z) - 4z = 0 \]This simplifies to \[ 3z + z - 4z = 0 \] which further simplifies to \[ 0 = 0 \]. This shows that the equation is consistent and doesn't provide additional conditions.

Step 5: Express Each Variable in terms of a Parameter

Since the system leads to consistent conditions, express each variable in terms of the parameter \( z \): \[ x = 3z, \quad y = -z, \quad z = z \]. Choose \( z \) as a parameter, say \( t \), then:\[ x = 3t, \quad y = -t, \quad z = t \].

Key Concepts

Substitution Method

The substitution method is a fundamental technique in solving systems of linear equations. First, identify a simple equation from the system where one variable can be isolated. In this case, the equation was \( y + z = 0 \). Here, \( y \) can easily be expressed in terms of \( z \) – resulting in \( y = -z \).
Then, substitute this expression into the other equations. For example, with \( y = -z \) substituted into the first equation \( x + y - 2z = 0 \), we obtain \( x - 3z = 0 \). This gives \( x = 3z \).
By isolating and replacing variables, we simplify the system. This method reduces the complexity by reducing the number of variables involved, making the equations easier to solve. The substitution step ties together the different pieces of the puzzle, allowing us to eventually find the expressions for all variables involved.

Parameterization

In many cases, like in our original problem, the system of equations does not yield distinct values for each variable. Instead, we express them in terms of a parameter.
After simplifying the problem using substitution, the expressions found for \( x \), \( y \), and \( z \) were \( x = 3z \), \( y = -z \), and \( z = z \).
Assign a parameter, say \( t \), which represents a free variable. You then express each variable in terms of this parameter:

• \( x = 3t \)
• \( y = -t \)
• \( z = t \)

Parameterization is useful for describing all possible solutions of a system that has infinitely many solutions, revealing the relationships between the variables under a consistent condition.

Linear Consistency

Linear consistency in a system of equations refers to whether the set of equations has at least one solution. In the provided example, following the application of the substitution method, the final test for consistency using \( 3z + z - 4z = 0 \) resulted in the identity \( 0 = 0 \).
This indicates not only the consistency of the system but also an indication that we have infinitely many solutions, as opposed to a unique solution or contradictory conditions causing no solution.
When a system shows such an identity, it opens the possibility for parameterization, as we’ve parameterized with \( t \). The consistency key ensures the system is solvable and guides us on how to find the complete solution set through parameterization.