Problem 7 - Precalculus Mathematics for Calculus

Question

Use back-substitution to solve the triangular system. $$\left\\{\begin{aligned} x-2 y+4 z &=3 \\ y+2 z &=7 \\ z &=2 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 3 \), \( z = 2 \).

Step 1: Identify the Known Variable

In the triangular system, we start with the equation where the variable is already isolated. Here, the equation is \( z = 2 \). Thus, the value of \( z \) is already known: \( z = 2 \).

Step 2: Substitute the Known Value into the Second Equation

Now that \( z = 2 \), substitute this value into the second equation: \( y + 2z = 7 \). This becomes \( y + 2(2) = 7 \) or \( y + 4 = 7 \).

Step 3: Solve for the Second Variable

From the equation \( y + 4 = 7 \), solve for \( y \) by subtracting 4 from both sides. This gives \( y = 7 - 4 \), so \( y = 3 \).

Step 4: Substitute Known Values into the First Equation

Using the values \( y = 3 \) and \( z = 2 \), substitute them into the first equation: \( x - 2y + 4z = 3 \). This becomes \( x - 2(3) + 4(2) = 3 \).

Step 5: Solve for the First Variable

Simplify the equation: \( x - 6 + 8 = 3 \), which simplifies to \( x + 2 = 3 \). By subtracting 2 from both sides, we find \( x = 1 \).

Key Concepts

Understanding Triangular Systems

A triangular system of linear equations is a special kind of system where the equations are organized in either an "upper triangular" or "lower triangular" format. This configuration makes solving the system much more straightforward using techniques like back-substitution. Let's look at how these systems are structured and why they are beneficial for solving linear equations.

• **Upper Triangular System:** In this setup, each equation involves fewer variables as you progress down the system. The last equation has only one variable, the second to last has two, and so on.
• **Lower Triangular System:** Conversely, this setup starts with simple equations involving fewer variables and increases the number as you move downwards. However, in the context of back-substitution, we focus on upper triangular systems.

Triangular systems are significant because they allow for a straightforward solution method – starting with the equation where the variable is already isolated and proceeding backward to find the other unknowns. This efficiency is crucial when it comes to computational methods for solving larger systems.

Solving Linear Equations

Linear equations are mathematical expressions that show a straight-line relationship between variables. Solving linear equations typically involves finding the value of unknowns that satisfy all given equations. In the context of triangular systems, the process is simplified by their structural arrangement.

To solve a linear system:

• Identify the equation where one of the variables is already known or can be easily found.
• Substitute this value into the other equations to reduce the number of unknowns.
• Continue substituting back into the previous equations until all variables are solved.

For instance, starting with the simplest equation in our triangular system, \[ z = 2 \], allows us to find specific values sequentially without complex calculations. This systematic unraveling of unknowns reflects the elegance of solving triangular systems.

The Step-by-Step Solution

Breaking down the process of solving a triangular system systematically not only makes it easier to follow but also reveals the efficiency behind such methods. Referring to the original problem, let's review each step in detail.

Step 1 begins with identifying the known variable, which in this case was found in the simplest equation, \[ z = 2 \]. Knowing this value lets us tackle equation 2 with more ease.

In Step 2, substituting \( z = 2 \) into \[ y + 2z = 7 \] converts the expression into a simpler form \[ y + 4 = 7 \], allowing for straightforward algebraic manipulation to solve for \( y \). This is solved by isolating \( y \) which leads to \( y = 3 \).

Finally, Step 3 solves for \( x \) by substituting both known values, \( y = 3 \) and \( z = 2 \), into the first equation \[ x - 2y + 4z = 3 \]. Simplifying this expression yields \( x = 1 \).

It's this organized and orderly method that streamlines solving triangular systems, moving from known quantities to uncover the unknowns efficiently.