91Ó°ÊÓ

When dealing with systems of linear equations, a **triangular system** is one in which each equation has fewer variables than the previous one. These types of systems are easy to solve because the equations are structured in such a way that you can solve them sequentially, from one end to the other. Triangular systems can either be **upper triangular systems** or **lower triangular systems**.

• In an upper triangular system, the equations are arranged such that each equation involves one less variable than the previous equation, proceeding from the first to last.
• A lower triangular system, conversely, arranges equations from the last to the first.

The given problem is a triangular system with the following structure:

• The first equation has two variables, x and z.
• The second equation involves y and z.
• The last equation is only in terms of z.

This structure allows us to begin solving the system with the simplest equation and work our way back to solve for all variables.

To solve a system of linear equations, we typically employ methods like **substitution**, **elimination**, or when applicable, utilize properties specific to the system structure, such as those in triangular systems. For our triangular system problem, a major technique that simplifies the process is **back-substitution**.

• **Back-substitution** involves solving for one variable at a time starting from the last equation in the system and "back-substituting" values into previous equations.
• This method is particularly effective and straightforward when working with triangular systems, which are already structured in a form that makes the process sequential.

Let's walk through the process in the context of our exercise: 1. **Start from the Bottom**: Solve the simplest equation first. In our system, the last equation is isolated for z. This gives us an immediate solution for z, which is then used in the preceding equations. 2. **Substitute and Solve**: Substitute the solved value into the next equation higher up (in this case, solve for y using the second equation). 3. **Repeat**: Continue this process until all variables are solved. The last step usually involves solving for the first variable (x in our exercise) using all known values for other variables.

The step-by-step solution provided in the original exercise helps clarify the back-substitution method and detailed solving of the triangular system. Let's breakdown these steps further for clear understanding: **Step 1: Solving for z** The equation \( \frac{1}{2}z = 4 \) requires simple multiplication to isolate \( z \). Once multiplied by 2, we find \( z = 8 \). **Step 2: Substitution for y** With \( z = 8 \), replace \( z \) in the second equation, \( 2y - z = -6 \). The equation simplifies to \( 2y = 2 \) after solving the arithmetic, leading to \( y = 1 \) when divided by 2. **Step 3: Substitution for x** Finally, knowing both \( z = 8 \) and \( y = 1 \), we substitute into the first equation \( 4x + 3z = 10 \). Substitute \( z = 8 \) into the equation to get \( 4x + 24 = 10 \). Simplify this to \( 4x = -14 \), and solving gives \( x = -\frac{7}{2} \). This methodical approach to solving ensures accuracy and completeness, providing clear foundational steps for mastering similar problems.