Problem 21 - Algebra and Trigonometry

Question

\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x &-4 z=1 \\ 2 x-y-6 z &=4 \\ 2 x+3 y-2 z &=8 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The complete solution is \((x, y, z) = (5, 0, 1)\).

Step 1: Write the system of equations

We have the following system of linear equations: 1. \(x - 4z = 1\) 2. \(2x - y - 6z = 4\) 3. \(2x + 3y - 2z = 8\)

Step 2: Express one variable in terms of others

From Equation (1): \(x - 4z = 1\) solve for \(x\):\[x = 1 + 4z\]

Step 3: Substitute into other equations

Substitute \(x = 1 + 4z\) into Equations (2) and (3):Equation (2):\(2(1+4z) - y - 6z = 4\) Simplifying gives:\[2 + 8z - y - 6z = 4 \Rightarrow -y + 2z = 2 \Rightarrow y = 2z - 2\]Equation (3):\(2(1+4z) + 3y - 2z = 8\)Simplifying gives:\[2 + 8z + 3y - 2z = 8 \Rightarrow 3y + 6z = 6 \Rightarrow y = -2z + 2\]

Step 4: Check consistency of the equations

Compare the expressions from Step 3 for \(y\):From Equation (2), \(y = 2z - 2\)From Equation (3), \(y = -2z + 2\)For these to hold true simultaneously -> \(2z - 2 = -2z + 2\).Solving gives: \[4z = 4\Rightarrow z = 1\]

Step 5: Find final values for x, y, and z

Now substitute \(z = 1\) back into expressions for \(x\) and \(y\):\(x = 1 + 4z = 1 + 4(1) = 5\)\(y = 2z - 2 = 2(1) - 2 = 0\) Thus, the solution is \((x, y, z) = (5, 0, 1)\).

Key Concepts

Linear Equations

Linear equations are the fundamental building blocks of algebra. They express relationships between quantities and involve variables that are raised only to the power of one. Each equation can be plotted as a straight line on a graph, hence the term "linear."
These equations frequently appear as simple structures like this:

• \( ax + by + cz = d \)

where \(a\), \(b\), and \(c\) are constants and \(x\), \(y\), and \(z\) are variables.
In our problem, we have a system of three linear equations, which we want to solve simultaneously to find the values of \(x\), \(y\), and \(z\). By working through the equations together, we can discover how these quantities relate to one another.

Substitution Method

The substitution method is a popular and effective approach to solving systems of linear equations.
Its primary goal is to express one variable in terms of another variable, making it easier to substitute later and find solutions.
Here’s how it works:

• First, select an equation and solve for one variable in terms of the others.
• Then, substitute this expression into the other equations, reducing the number of variables and simplifying the problem.

In our problem, we chose the first equation, \(x - 4z = 1\), and rearranged it to \(x = 1 + 4z\).
By substituting this expression into the remaining two equations, we reduced the complexity and focused on finding \(y\) and \(z\). This step is crucial as it sets the stage for solving the entire system.

Consistency of Equations

The consistency of equations is determined by examining whether the equations can all be satisfied by the same set of variable values.
A consistent system has at least one solution, while an inconsistent system has no solution, usually due to contradictory equations.
In a linear system, consistency can often be checked by substituting solutions and checking for any contradictions.

• For example, after substituting \(x = 1 + 4z\) into equations, we compared different expressions obtained for \(y\).
• In this exercise, we found \(y = 2z - 2\) from one equation and \(y = -2z + 2\) from another.

Equating these gave us \(4z = 4\), leading to \(z = 1\). Thus, the system is consistent, as there is a common solution satisfying all equations simultaneously.

Solving for Variables

Once consistency is established, the next step involves solving for the individual variable values.
Through earlier substitutions, we narrowed down our focus to specific expressions for each variable.
Here’s a step-by-step breakdown:

• We found \(z = 1\) from our consistency check.
• Substituting \(z = 1\) back into the expression for \(y = 2z - 2\) gives \(y = 0\).
• Finally, substituting \(z = 1\) in \(x = 1 + 4z\), we find \(x = 5\).

Thus, the final solution for the system of equations is \((x, y, z) = (5, 0, 1)\).
By following this logical process, complex systems of equations can be tackled with clarity and confidence.