Problem 10 - Precalculus: Mathematics for Calculus

Question

Use back-substitution to solve the triangular system. \(\left\\{\begin{aligned} x-2 y+3 z &=10 \\ 2 y-z &=2 \\ 3 z &=12 \end{aligned}\right.\)

Step-by-Step Solution

Answer

x = 4, y = 3, z = 4.

Step 1: Solve for z from the last equation

The last equation is \( 3z = 12 \). To solve for \( z \), divide both sides by 3: \[ z = \frac{12}{3} = 4 \].

Step 2: Use the value of z to solve for y in the second equation

The second equation is \( 2y - z = 2 \). Substitute \( z = 4 \) into the equation: \[ 2y - 4 = 2 \]. Then, add 4 to both sides: \[ 2y = 6 \]. Finally, divide both sides by 2 to find \( y \): \[ y = \frac{6}{2} = 3 \].

Step 3: Use the values of y and z to solve for x in the first equation

The first equation is \( x - 2y + 3z = 10 \). Substitute \( y = 3 \) and \( z = 4 \) into the equation: \[ x - 2(3) + 3(4) = 10 \]. This simplifies to \[ x - 6 + 12 = 10 \]. Combine the constants: \[ x + 6 = 10 \]. Subtract 6 from both sides to solve for \( x \): \[ x = 4 \].

Key Concepts

Triangular System

A triangular system of linear equations is a special form of a system where the equations are arranged in a way that makes solving them simpler.
Specifically, it means that each subsequent equation in the system contains fewer variables than the previous one. This results in the last equation having only one variable, the second last having two, and so forth.
In a triangular system, you typically start by solving the simplest equation first, and then use that solution to progressively solve the other equations. To illustrate, consider the given system:

• The last equation, \(3z = 12\), features only \(z\).
• The second equation, \(2y - z = 2\), involves \(y\) and \(z\).
• The first equation, \(x - 2y + 3z = 10\), includes \(x\), \(y\), and \(z\).

This hierarchical structure allows for easier application of back-substitution, enabling us to solve each variable step by step.

Linear Equations

Linear equations are expressions where the variables are only raised to the first power, which means they form straight lines when graphed.
These equations represent relationships with a constant rate of change and are fundamental to understanding systems of equations.

When dealing with linear equations, each equation defines a line on a coordinate plane, with solutions represented by the intersection points of these lines. In a system of linear equations, you are essentially searching for such intersection points, which denote the values for each variable that satisfy all equations simultaneously.For example, our system:

• \(x - 2y + 3z = 10\)
• \(2y - z = 2\)
• \(3z = 12\)

Utilizes these principles. Each solution found through the process of back-substitution indicates where these linear equations intersect in a three-dimensional space.

Solving Systems of Equations

Solving systems of equations involves finding the set of values for the variables that satisfy all equations in the system simultaneously.
There are various methods available, but for a triangular system, back substitution is particularly efficient.

The process starts with the simplest equation, which is usually at the bottom of a triangular system and moves upwards, substituting known values to reduce the complexity of each equation.For our example:

• We first solved \(3z = 12\), giving us \(z = 4\).
• Next, using \(z = 4\), we solved \(2y - z = 2\) to find \(y = 3\).
• Finally, with both \(y = 3\) and \(z = 4\), we found \(x = 4\) by solving \(x - 2y + 3z = 10\).

Back substitution leverages the structure of a triangular system to simplify the computation, making it a powerful method for resolving linear systems efficiently.