Problem 8 - Algebra and Trigonometry

Question

Use back-substitution to solve the triangular system. $$ \left\\{\begin{aligned} x-2 y+3 z &=10 \\ 2 y-z &=2 \\ 3 z &=12 \end{aligned}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Solve for the last variable

Start with the equation that has only one variable, which is the third equation. It says:\[ 3z = 12 \]To solve for \( z \), divide both sides by 3:\[ z = \frac{12}{3} = 4 \].

Step 2: Substitute known value into the second equation

Now that we know \( z = 4 \), substitute this value into the second equation:\[ 2y - z = 2 \]Replace \( z \) with 4:\[ 2y - 4 = 2 \]Add 4 to both sides to solve for \( y \):\[ 2y = 6 \]Divide both sides by 2:\[ y = \frac{6}{2} = 3 \].

Step 3: Substitute known values into the first equation

Now substitute \( y = 3 \) and \( z = 4 \) into the first equation:\[ x - 2y + 3z = 10 \]Replace \( y \) and \( z \):\[ x - 2(3) + 3(4) = 10 \]Simplify:\[ x - 6 + 12 = 10 \]Combine like terms:\[ x + 6 = 10 \]Subtract 6 from both sides to solve for \( x \):\[ x = 10 - 6 = 4 \].

Key Concepts

Back-Substitution

Back-substitution is a clever method used to solve triangular systems of equations. Imagine having equations piled like a staircase, each dependent on the one just below it. This technique starts at the bottom, where the simplest equation lies, and solves it first. Once the bottom variable is determined, you climb upwards.
If an equation contains more than one unknown, plug in the values you've already found until all the unknowns are solved. This step-by-step approach is excellent for systems transformed into triangular form. It saves you time and frustration, ensuring accuracy along the way.

Linear Equations

Linear equations are the foundational blocks of algebra, describing a straight line when plotted on a graph. In these types of equations, each power of the variable is 1, making them simple and direct.
A typical linear equation appears as either a simple equation like \( ax + b = c \) or as a system of equations like our triangular system. These systems can be solved using methods such as substitution, elimination, or matrices when more equations are involved.

• Linear equations can have one solution, no solution, or infinitely many solutions.
• Being able to recognize and solve linear equations helps in deciphering more complex algebraic expressions.

Algebraic Solution Steps

Solving any equation involves a process—known as the algebraic solution steps. These steps ensure that every action is logical and leads to finding the correct answer.
Let's break down the solution steps:

• Simplification: If needed, simplify each equation by performing operations like combining like terms.
• Solving for a Variable: Focus on one equation and solve for a variable. This frequently involves isolating the variable by adding, subtracting, multiplying, or dividing what's necessary on both sides.
• Substitution: With known variables, substitute their values into the remaining equations. Repeat the solving process if required, to find all unknowns.
• Verification: Once all variables are found, plug them back into the original equations to verify your solution.

Following these structured steps ensures clarity and correctness, making the solutions of even complex systems simpler.