Problem 5 - Precalculus: Mathematical 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: Recognize the Form of the System

The system is triangular, meaning it is already in a form that allows us to solve each equation starting from the bottom and working upwards.

Step 1: Solve for the First Variable

The last equation is \( z = 2 \). Since it is already solved for \( z \), we have our first solution: \( z = 2 \).

Step 2: Substitute and Solve for the Second Variable

Substitute \( z = 2 \) into the second equation: \( y + 2(2) = 7 \), which simplifies to \( y + 4 = 7 \). Solving this, we subtract 4 from both sides to get \( y = 3 \).

Step 3: Substitute and Solve for the Third Variable

Substitute \( y = 3 \) and \( z = 2 \) into the first equation: \( x - 2(3) + 4(2) = 3 \). Simplify to \( x - 6 + 8 = 3 \). Combine the constants to get \( x + 2 = 3 \). Then subtract 2 from both sides to find \( x = 1 \).

Step 5: Verify the Solution

Substitute \( x = 1 \), \( y = 3 \), and \( z = 2 \) back into all original equations to confirm they all hold true, which they do.

Key Concepts

Understanding Triangular Systems

A triangular system is a specific arrangement of equations that makes solving them straightforward, especially when employing back-substitution. In mathematics, a system of equations is triangular when each equation, from top to bottom, adds one more unknown than the previous one and is arranged from the simplest to the most complex. Here, the system is already in a lower triangular form:

• The last equation incorporates only the unknown \( z \).
• The second equation involves \( y \) and \( z \).
• The first equation contains \( x \), \( y \), and \( z \).

This setup allows for efficient resolution by starting at the simplest equation (bottom) and substituting upward. This is because the unknowns are progressively revealed, simplifying the process as each previous solution is substituted into the subsequent equation.

Solving Equations with Back-Substitution

Solving equations in a triangular system through back-substitution is an efficient method because it leverages the structure of the system. Here’s how it unfolds:

• **Start with the last equation:** Since it contains only one unknown, solve it directly. In our case, \( z = 2 \), giving us a clear starting point.
• **Substitute upwards:** Use the obtained value of \( z \) in the second equation. Substitute \( z = 2 \) into \( y + 2z = 7 \) to find \( y = 3 \).
• **Continue substituting upwards:** Use the solutions of \( y \) and \( z \) in the first equation \( x - 2y + 4z = 3 \) to find \( x = 1 \).

This method ensures that each unknown is solved in a sequence that builds on previous results, making it systematic and easy to follow.

Algebra Steps in Back-Substitution

The algebra steps involved in back-substitution are simple yet critical in reaching a solution accurately. Each solution to an equation depends on carefully performing arithmetic operations that maintain equilibrium across the equation.

• Direct substitution: In the second step, substitute \( z = 2 \) into the second equation and simplify: \( y + 2 \times 2 = 7 \) results in \( y = 3 \).
• Simplification: For each step, ensure you simplify every expression. For example, simplifying \( x - 6 + 8 = 3 \) yields \( x + 2 = 3 \).
• Verification: Finally, substitute all found values back into the original equations to confirm their correctness. If done accurately, the equations should hold true, as they do for \( x = 1 \), \( y = 3 \), \( z = 2 \).

Engage in these algebraic steps rigorously to achieve reliable results in any triangular system.