Problem 6 - Precalculus: Mathematical for Calculus

Question

Use back-substitution to solve the triangular system. $$\left\\{\begin{aligned} x+y-3 z &=8 \\ y-3 z &=5 \\ z &=-1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Solve for the Bottom Variable

We start with the simplest equation from the system, which is the bottom one: \[ z = -1 \] This equation clearly states the value of \( z \). There are no further simplifications needed.

Step 2: Substitute and Solve for the Middle Variable

Now, we substitute \( z = -1 \) into the second equation:\[ y - 3(-1) = 5 \]This simplifies to:\[ y + 3 = 5 \]Now, solve for \( y \):\[ y = 5 - 3 = 2 \]

Step 3: Substitute and Solve for the Top Variable

Finally, substitute \( y = 2 \) and \( z = -1 \) into the first equation:\[ x + 2 - 3(-1) = 8 \]Simplify the equation:\[ x + 2 + 3 = 8 \]\[ x + 5 = 8 \]Solve for \( x \):\[ x = 8 - 5 = 3 \]

Key Concepts

Triangular System

A triangular system in algebra is an arrangement of equations that simplifies solving for variables in a systematic way. Specifically, a system is called triangular if it resembles a step-like formation in terms of equations and variables. Each equation in the system generally involves fewer variables than the previous one.
Triangular systems are particularly useful because they allow the process of back-substitution. Back-substitution starts from the bottom of the system, where a simple equation gives the value for one variable. This known value can then be substituted into the equations above, making it easier to find the remaining variables.
In our given problem, the triangular system is already set up so that it is easy to solve using back-substitution. The bottom equation is the simplest and provides the value for one of the variables immediately.

Algebraic Equations

Algebraic equations form the backbone of solving mathematical problems involving unknowns. These are equations composed of variables, coefficients, and constants connected by algebraic operations such as addition, subtraction, multiplication, and division.
An essential component of understanding algebraic equations is knowing their structure and the role of each part.

• Variables: symbols representing unknown quantities. For example, in our system, \(x, y,\) and \(z\) are variables.
• Coefficients: numerical factors multiplying the variables. In the equation \(x + y - 3z = 8\), the coefficients are 1, 1, and -3 respectively.
• Constants: stand-alone numbers that do not multiply with variables. Here, 8, 5, and -1 are constants in different equations.

Algebraic equations can describe various relationships and are adaptable for different mathematical applications. Solving them often involves rearranging and simplifying them to find the values of the unknown variables.

Solving Systems of Equations

Solving systems of equations involves finding values for variables that satisfy all equations simultaneously. In a system, each equation represents a constraint on the values that variables can take.
There are several methods to solve systems of equations, with back-substitution being one of the most straightforward for triangular systems. The step-by-step approach allows for logical progression:

• Start from the simplest equation, typically located at the bottom of the triangular system.
• Find the value of one variable, which then helps in solving other equations.
• Proceed upwards, substituting known values to find remaining unknowns.

In our example, once you solve for \(z\), this value is used to solve the equation for \(y\). Finally, knowing \(y\) and \(z\), you can solve for \(x\). This systematic approach ensures that all equations are satisfied simultaneously, providing a consistent solution to the system.