Problem 32 - Algebra and Trigonometry

Question

25–34 Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$\left\\{\begin{aligned} 3 r+2 s-3 t &=10 \\ r-s-t &=-5 \\ r+4 s-t &=20 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system is dependent with solution: \( (t, 5, t) \) for all real \( t \).

Step 1: Write the System of Equations

Start with the given system of equations:\[ \begin{align*} 1) &\quad 3r + 2s - 3t = 10 \ 2) &\quad r - s - t = -5 \ 3) &\quad r + 4s - t = 20 \end{align*} \]

Step 2: Eliminate a Variable

Subtract equation (2) from equation (3) to eliminate \(r\):\[ (r + 4s - t) - (r - s - t) = 20 - (-5) \]Simplifying gives:\[ 5s = 25 \]\[ s = 5 \]

Step 3: Substitute Back into Equations

Substitute \(s = 5\) into equation (2):\[ r - 5 - t = -5 \]Simplifying gives:\[ r - t = 0 \]So, \( r = t \).

Step 4: Back-Substitute to Find Complete Solution

Substitute \(r = t\) and \(s = 5\) into equation (1):\[ 3t + 2(5) - 3t = 10 \]\[ 10 = 10 \]This equation holds true for all \(t\), confirming that the system is dependent. The solution is:\[ \{ (t, 5, t) | t \in \mathbb{R} \} \]

Step 5: Conclusion

The system of equations is dependent with the solution as parametric form \( (t, 5, t) \) for all real \(t\).

Key Concepts

Linear Equations

Linear equations are equations in which each term is either a constant or the product of a constant and a single variable. These equations don't involve exponents or powers beyond one, making them straight lines when graphed on a coordinate plane. In the context of multiple linear equations, you often solve for the intersection points, which provide common solutions for all equations in the system.

Key features of linear equations include:

• Simplicity in form: They have basic terms like \(ax + by = c\).
• Pictorial representation: Their graph is a straight line.
• Applications: Used in various fields for modeling and problem-solving.

In solving linear systems, understanding how the equations interact helps identify if they are dependent, independent, or inconsistent. Dependent systems have infinite solutions because their lines overlap entirely.

Parametric Solutions

A parametric solution expresses the infinite solutions of a dependent system using a free variable, often denoted by a parameter like \(t\). This formalizes the redundancy in equations where more than one equation provides the same information mathematically.

For instance, in the given exercise, the system of equations simplifies to having one or more overlapping solutions. After reducing the original equations, the solution was given by \( (t, 5, t) \), showing the degrees of freedom with the parameter \(t\). Here, the parameter \(t\) represents any real number, showing infinite possibilities in solutions.

Parametric solutions highlight:

• Flexibility: You describe the solution space in terms of one or more parameters.
• Completeness: They encapsulate all possible solutions efficiently.

Use parametric solutions to simplify situations where solving directly might be overly complex or redundant.

Variable Elimination

Variable elimination is a method to simplify systems of equations by strategically removing (eliminating) certain variables to make solving easier. This technique typically involves adding or subtracting equations from each other to remove a variable, thus reducing the complexity of the system.

In the provided exercise, variable elimination was used to simplify the system. By subtracting the second equation from the third, the variable \(r\) was successfully eliminated, resulting in a simpler equation \(5s = 25\), which was straightforward to solve for \(s\).

Important points about variable elimination:

• Efficiency: Quickly reduces the number of variables, making systems manageable.
• Sequential: Often done in steps, simplifying one variable at a time until the system is resolved.
• Versatility: Can be used alongside substitution and graphing methods as part of a combined approach.

Mastering variable elimination aids in solving linear systems effectively, paving the way for deeper mathematical problem-solving.