Problem 7 - Algebra and Trigonometry

Question

\(7-12=\) 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: Solve for the last variable z

The last equation in the system is already solved for z. It states that \( z = 2 \). This means we have found the value of \( z \) to be 2.

Step 2: Substitute z into the second equation to find y

Take the value of \( z = 2 \) and substitute it into the second equation, \( y + 2z = 7 \). This gives us \( y + 2(2) = 7 \). Simplifying, we get \( y + 4 = 7 \). Solving for \( y \), we subtract 4 from both sides to find \( y = 3 \).

Step 3: Substitute y and z into the first equation to find x

Now that we know \( y = 3 \) and \( z = 2 \), substitute these values into the first equation: \( x - 2y + 4z = 3 \). This becomes \( x - 2(3) + 4(2) = 3 \). Simplifying, we get \( x - 6 + 8 = 3 \), which simplifies further to \( x + 2 = 3 \). Solving for \( x \) by subtracting 2 from both sides, we find \( x = 1 \).

Key Concepts

Triangular System

A triangular system is a specific type of linear equation system, where the arrangement of equations forms a structure similar to a triangle. In our original exercise, we are given three equations that fit this criterion. The initial equation involves three variables, the second has two, and the last equation only contains one variable.
This 'ladder-like' structure makes it much easier to apply back-substitution, simplifying the process of solving the system of equations one step at a time.
Key characteristics of a triangular system include:

• The presence of a 'triangular' form, where each subsequent equation contains one fewer variable than the previous.
• A natural sequence for solving the variables, often starting from the last equation upward to the first.
• The upper or lower triangular form, crucial for efficient backward or forward substitution methods respectively.

Understanding triangular systems assists in breaking down complex algebraic problems into more manageable parts.

Algebraic Equations

Algebraic equations form the backbone of solving any mathematical model. They involve equalities formed by algebraic expressions. In the context of our triangular system, each line in our system is an individual algebraic equation that collectively defines the system.
Typically, these equations are presented in a standard form where each variable and constant is clearly delineated, as such:

• The variable terms (like 'x', 'y', and 'z') represent the unknowns you need to solve.
• The coefficients attached to these variables (numbers preceding the variables) indicate how each unknown relates to one another.
• Constant terms (like 3, 7, and 2 in the original problem) define the total value that the expression equates to.

Understanding these equations allows you to manipulate and solve them, especially within systems where multiple variables interact.

Solving Systems of Equations

Solving systems of equations can be approached using a variety of methods, with back-substitution being one of the most straightforward techniques for triangular systems. In our case, we begin with the simplest equation and solve for one variable. Here’s how back-substitution works:

• **Begin with the last equation**: It typically contains the fewest variables. Solve for the variable outright, as this gives you a concrete value to work with.
• **Substitute backward through preceding equations**: Use the value from the solved equation to simplify the prior one, reducing the number of unknowns step by step.
• **Continue until all variables are solved**: By consistently feeding the known values back into the earlier equations, you ensure that each step simplifies the unknowns effectively, ultimately solving the entire system.

This method reduces complexity and minimizes error, making the process of tackling multiple algebraic equations methodical and systematic.