Problem 7 - Algebra and Trigonometry

Question

Use back-substitution to solve the triangular system. $$ \left\\{\begin{aligned} x+2 y+z &=7 \\\\-y+3 z &=9 \\ 2 z &=6 \end{aligned}\right. $$

Step-by-Step Solution

Answer

\(x = 4\), \(y = 0\), \(z = 3\)

Step 1: Solve the Last Equation

The last equation is simple and in the form of a single variable equation: \[ 2z = 6 \]To find \(z\), divide both sides by 2:\[ z = \frac{6}{2} = 3 \]Thus, \(z = 3\).

Step 2: Substitute z into the Second Equation

Now that we have \(z = 3\), substitute this value into the second equation:\[-y + 3z = 9\]Substitute \(z = 3\):\[-y + 3(3) = 9\]This simplifies to:\[-y + 9 = 9\]Subtract 9 from both sides:\[-y = 0\]Multiply by -1:\(y = 0\)Therefore, \(y = 0\).

Step 3: Substitute y and z into the First Equation

With \(y = 0\) and \(z = 3\), substitute these values into the first equation:\[x + 2y + z = 7\]Substitute \(y = 0\) and \(z = 3\):\[x + 2(0) + 3 = 7\]This simplifies to:\[x + 3 = 7\]Subtract 3 from both sides:\[x = 4\]Thus, \(x = 4\).

Key Concepts

Back-Substitution

Back-substitution is a method used to solve linear equations, particularly when dealing with a triangular system. This approach is especially handy in a set of equations that are structured to easily solve one variable at a time. In back-substitution, we start solving from the equation with the fewest variables, usually the last one in a triangular system.
First, find the value of the last unknown variable. Once solved, substitute this value back into the previous equations. This makes solving the other variables a sequential process.

• Begin with the simplest equation, typically involving one variable.
• Solve for that variable.
• Substitute the obtained value into preceding equations.
• Continue this method systematically upwards through the system.

By following these steps, you break down complex problems into simpler parts. This makes the solution more manageable and easier to understand.

Triangular System

A triangular system is a set of linear equations where each equation has a successively larger number of terms. In an upper triangular system, the first equation might involve all variables, the second equation one less, and so on until the last equation, which involves a single variable. Such systems are designed for solving easily with back-substitution.
This structured form simplifies the process of finding solutions by progressive elimination of variables. - **Upper Triangular Form**: - The coefficients below the diagonal are zero. - It appears like a triangle with the non-zero elements above (or) the zero elements below the diagonal. Understanding the concept of a triangular system helps to visualize the equation arrangement. This visualization assists in logically applying the back-substitution technique for solving.

Algebraic Solutions

Finding algebraic solutions involves determining the variable values in the equations using algebraic manipulation. The goal is to express the variables accordingly, often starting with the simplest equation to isolate one variable and substitute it throughout the system.
This process often consists of:

• Isolating a variable in one equation.
• Substituting this variable in other equations to reduce the complexity.
• Continuing this substitution until all variables are found.

Using algebraic solutions to solve systems of equations ensures a precise analytical approach. With practice, these techniques become second nature, helping to tackle even more complex systems with confidence.