Problem 9 - Algebra and Trigonometry

Question

\(7-12=\) 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

The solution is \(x = 4\), \(y = 0\), \(z = 3\).

Step 1: Solve for z

Start with the last equation since it's the simplest: \(2z = 6\). Divide both sides by 2 to solve for \(z\): \[ z = \frac{6}{2} = 3 \] Thus, we have found that \( z = 3 \).

Step 2: Back-substitution into second equation

Now substitute the value of \(z\) into the second equation: \(-y + 3(3) = 9\). This simplifies to: \[-y + 9 = 9\] Subtract 9 from both sides: \[-y = 0\] Thus, \( y = 0 \).

Step 3: Back-substitution into first equation

Substitute the values of \(y\) and \(z\) into the first equation: \( x + 2(0) + 3 = 7 \). Simplify: \[ x + 3 = 7 \] Subtract 3 from both sides: \[ x = 4 \] Therefore, \( x = 4 \).

Step 4: Verify solution

Verify by substituting \(x = 4\), \(y = 0\), and \(z = 3\) back into the original equations. First equation: \(4 + 2(0) + 3 = 7\), which simplifies to \(7 = 7\). Second equation: \(-0 + 3(3) = 9\), which simplifies to \(9 = 9\). Third equation: \(2(3) = 6\), which simplifies to \(6 = 6\). All equations hold true, confirming our solution is correct.

Key Concepts

Back-substitution

Back-substitution is a powerful technique for solving systems of linear equations, particularly in a triangular format. A triangular system of equations is one where each equation has fewer variables than the previous one. This method is highly efficient and logical. You start solving from the bottom of the system, where usually the simplest equation resides, and work your way upwards. As you solve for one variable, you substitute its value back into previous equations to find other unknowns. This step-by-step approach ensures that you use already solved values to ease the process of finding remaining unknowns. In the given problem, we start with solving for the last variable using the simplest equation, and then back-substitute the found values to solve the preceding ones. It's crucial to follow each step meticulously, ensuring accuracy in each calculation.

Solving Linear Equations

Solving linear equations involves finding the values of unknown variables that make the equation true. These equations are called 'linear' because they form a straight line when graphed on a coordinate plane. A linear equation usually follows the structure of an algebraic expression set equal to something. Solving these equations requires some basic algebraic manipulations, such as addition, subtraction, multiplication, or division. In our example, each equation is progressively solved using back-substitution.

• First, find the value of the easiest variable in the simplest equation.
• Then, substitute this value in the next equation to determine another variable.
• Continue this process until all variables are found.

Each step relies on the consistency and correctness of the previous steps, highlighting the importance of solving in the right order to maintain validity.

Systems of Equations

Systems of equations are sets of equations that share variables. The goal is to find a specific set of variable values that satisfy all equations simultaneously. These are commonly represented as a set of equations, as shown in the original exercise, where there are three equations with three variables. Different methods, such as substitution, elimination, and graphical methods, can be used to solve these systems. In this exercise problem, a triangular system allows for a more straightforward solving process through back-substitution. It is important to always ensure that each equation is consistent within the larger system. When you find values that work for one equation, they must also apply to all others in the system to be valid. Confirming your solutions by substituting back into the original system is essential in verifying the correctness of your results.