Problem 54 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{aligned} 3 x+y &=2 \\ -4 x+3 y+z &=4 \\ 2 x+5 y+z &=0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Parameterized solution: \( x = t, y = 2-3t, z = 13t-2 \).

Step 1: Choose Variables to Eliminate

We start by choosing a variable to eliminate from at least two of the equations. Here, we will eliminate \( z \) by subtracting Equation 2 from Equation 3 to form a new equation.

Step 2: Eliminate \( z \) from Equations 2 and 3

Subtract Equation 2 from Equation 3: \[(2x + 5y + z) - (-4x + 3y + z) = 0 - 4\]Simplify to get:\[6x + 2y = -4 \]This is our new Equation 4.

Step 3: Solve for One Variable

Next, solve for \( x \) using Equation 1 and Equation 4.Equation 1 is \( 3x + y = 2 \).Solve Equation 4 for \( y \):\[y = -3x - 2 \]Now substitute this into Equation 4.

Step 4: Substitute and Simplify

Substitute \( y = -3x - 2 \) back into Equation 4:\[6x + 2(-3x - 2) = -4\]Simplify:\[6x - 6x - 4 = -4\]\[0 = 0 \]This means the substituted equation is consistent, suggesting a dependency between equations.

Step 5: Express Other Variables in Terms of Free Variable

Since Equation 4 is a zero equation, we return to the original equations and express all variables in terms of a parameter. Choose \( t = x \):From Equation 1:\[y = 2 - 3t\]Substitute \( y \) into Equation 2 or 3 to solve for \( z \):\[-4t + 3(2 - 3t) + z = 4 \]\[-4t + 6 - 9t + z = 4\]\[z = 13t - 2\].

Step 6: Write the General Solution

Write the general solutions for each variable using the parameter \( t \):\[x = t\]\[y = 2 - 3t\]\[z = 13t - 2\].

Key Concepts

Elimination Method

The Elimination Method is a strategic approach used to solve systems of linear equations by removing variables to simplify the problem and eventually find the values of the remaining variables. This method often involves adding or subtracting equations to cancel out a variable, which simplifies the system of equations.
In the given problem, our primary goal was to eliminate the variable \(z\) from the second and third equations. By subtracting the second equation from the third, we successfully removed \(z\), resulting in a simpler equation: \(6x + 2y = -4\). This new equation, alongside the first equation \(3x + y = 2\), allows us to focus solely on \(x\) and \(y\), effectively reducing the complexity of the original system.
To effectively use the elimination method, it's crucial to:

• Carefully choose the variable you want to eliminate. Often, this is based on coefficients that can be quickly modified for elimination.
• Ensure correct arithmetic operations to avoid mistakes.
• Simplify equations after each step to make calculations manageable.

This method simplifies solving complex systems step-by-step, ushering toward the ultimate solutions.

Systems of Equations

Systems of Equations are a collection of two or more equations with a shared set of variables. Solving them means finding the values of these variables that satisfy all the equations simultaneously.
In our exercise, the system involves three equations and three variables \(x, y, z\):

• \(3x + y = 2\)
• \(-4x + 3y + z = 4\)
• \(2x + 5y + z = 0\)

Such systems can have one unique solution, no solution, or infinitely many solutions. By using the elimination method, we transformed the system into one where variables depend on parameters, leading to infinitely many solutions.
Understanding systems of equations:

• Linear equations graph as straight lines in two-dimensional systems, and intersecting lines usually indicate a single solution.
• Parallel lines describe no solution scenarios, as they never meet.
• Overlapping lines lead to infinite solutions, as they share every point.

Recognizing these outcomes helps in predicting the nature of solutions when facing more complicated systems.

Parameterization

Parameterization occurs when an infinite number of solutions exist and the solutions are expressed in terms of one or more parameters. In this context, a parameter is a free variable that defines the solution set. This approach is typically employed when the system’s equations are not independent, hinting at potential infinite solutions.
For our example, once the elimination process simplified our equations, we ended up with a dependency between them. Recognizing this, we introduced a parameter \(t\), dictated by the choice \(t = x\). Consequently, we expressed all variables in terms of \(t\):

• \(x = t\)
• \(y = 2 - 3t\)
• \(z = 13t - 2\)

When parameterization is applied, each value of \(t\) yields a simultaneous solution for \(x, y, z\), crafting a continuum of solutions along a line in this context. It provides a comprehensive depiction of the solution set, ensuring all possibilities are considered. This technique effectively articulates the broad scope of solutions present in dependent systems.