Problem 8 - Algebra and Trigonometry

Question

\(7-12=\) Use back-substitution to solve the triangular system. $$ \left\\{\begin{aligned} x+y-3 z &=8 \\ y-3 z &=5 \\ z &=-1 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 3 \), \( y = 2 \), \( z = -1 \).

Step 1: Solve for the innermost equation

Starting from the last equation, solve for the variable directly given. The last equation is \( z = -1 \).

Step 2: Substitute the value of z into the second equation

The second equation is \( y - 3z = 5 \). Substitute \( z = -1 \) into the equation: \[ y - 3(-1) = 5 \]. This simplifies to \( y + 3 = 5 \).

Step 3: Solve for y

From the equation \( y + 3 = 5 \), solve for \( y \) by subtracting 3 from both sides: \( y = 2 \).

Step 4: Substitute the values of y and z into the first equation

Now that we have \( y = 2 \) and \( z = -1 \), substitute these values into the first equation: \( x + y - 3z = 8 \), which becomes \( x + 2 - 3(-1) = 8 \).

Step 5: Solve for x

Simplify the equation from the previous step: \( x + 2 + 3 = 8 \). Combine like terms to get \( x + 5 = 8 \). Solve for \( x \) by subtracting 5 from both sides: \( x = 3 \).

Key Concepts

Back-Substitution

Back-substitution is a method often used in linear algebra to solve a system of linear equations. It's particularly useful when you're dealing with a triangular system of equations. Think of it like peeling the layers of an onion—starting from the innermost equation (or the bottom of the triangle) and working your way up.
In this exercise, we had a set of three equations. We started by solving the simplest one: \( z = -1 \). Once we have determined the value of \( z \), we can substitute this value upwards through the other equations. This lets us detangle the variables one by one in a systematic manner. It's a sequential process:

• First, find the value of the last variable.
• Substitute it into the preceding equation to find another variable.


• Repeat until all variables are solved.

This method is highly reliable when you have a neatly structured triangular system, which is basically what ensures efficiency since we start from a known value and substitute it upwards.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. In the realm of solving linear equations, algebra helps us to represent problems in a generalized form and then find unknown values.
In this exercise, using algebra was key to simplifying and rearranging each equation to find one variable at a time. We use basic algebraic operations such as:

• Addition and subtraction to shift terms from one side of the equation to another.


• Substitution to replace variables with known values.
• Simplification to consolidate terms and make computations straightforward.

These techniques are the building blocks for solving more complex mathematical problems. Mastering algebraic manipulations allows for a systematic approach to problem-solving, which in this case, enabled us to find values for \( x\), \( y\), and \( z\) effectively.

Linear Equations

Linear equations are mathematical statements of equality that involve variables raised to the power of one. They're called 'linear' because they graph as straight lines when plotted on a coordinate plane.
In this problem, we dealt with a system of linear equations organized in a triangular format. Each equation involves one or more variables,

• The simplest form was \( z = -1 \), a single-variable equation which is straightforward to solve.


• The next layer was \( y - 3z = 5 \), which became simpler after substituting the known value of \( z \).
• Finally, the equation \( x + y - 3z = 8 \) required substitution of both \( y\) and \( z \) to solve for \( x \).

Systems of linear equations can model real-world situations and solve practical problems. Understanding them is pivotal because they form the basis for more advanced mathematical concepts such as matrices and vector spaces.