Problem 13 - Precalculus with Limits

Question

Using Back-Substitution In Exercises \(11 - 16 ,\) use back-substitution to solve the system of linear equations. $$\left\\{ \begin{aligned} 2 x + y - 3 z & = 10 \\ y + z & = 12 \\ z & = 2 \end{aligned} \right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 3\), \(y = 10\), and \(z = 2\).

Step 1: Identify the variable to substitute

We have three linear equations and the third equation gives us the value of \( z \) which is 2. As the method requests, we will substitute this value into the higher equations.

Step 2: Substitute z in equation 2

Substituting \( z = 2 \) into the second equation \(y + z = 12\), you get a new equation: \(y = 12 - 2 = 10\). Now we've solved for \( y \) as well.

Step 3: Substitute y and z in equation 1

Substituting \( y = 10 \) and \( z = 2 \) into first equation \(2x + y - 3z = 10\), we get a new equation: \(2x + 10 - 3*2 = 10\), then simplifying gives \(2x + 10 - 6 = 10\), this reduces to \(2x = 10 - 10 + 6\), this simplifies to \(2x = 6\), which reduces further to \(x = 3\). Now we've solved for \( x \) as well.

Step 4: Interpret the result

The interpretation of these results is that for this system of linear equations, the solutions are \(x = 3\), \(y = 10\), and \(z = 2\). This is your final answer.

Key Concepts

System of Linear Equations

In algebra, a system of linear equations consists of two or more linear equations made up of two or more variables, so that each equation is linear in each variable. A common way to solve these systems is by finding values for the variables that satisfy all equations simultaneously.

A simple example is a system with three equations and three variables, like the one in our exercise:

1. \(2x + y - 3z = 10\)
2. \(y + z = 12\)
3. \(z = 2\)

The objective is to determine the values of the variables \(x\), \(y\), and \(z\) that make all three equations true at the same time.

Systems of linear equations can be solved using various methods such as graphing, substitution, elimination, and matrix operations. The choice of method often depends on the complexity of the system and the preferences of the person solving the system.

Algebraic Substitution

The method of algebraic substitution is a technique used to solve systems of linear equations. It involves replacing variables with known values or expressions in order to simplify the equation and solve for the unknowns.

The process typically starts with the simplest equation that can easily be solved for one variable. In our exercise, the third equation \(z = 2\) already gives us the value of \(z\). This known value is then substituted into other equations in the system to find the remaining variables.

For instance, in the second step of our solution, we took the known value of \(z\) and substituted it into the second equation to find the value of \(y\). This process of substitution continues until all variables have been determined. Substitution is a powerful strategy when dealing with systems where one equation is significantly simpler than the others or when direct values for a variable are provided.

Solving Linear Systems

When it comes to solving linear systems, there are several methods available, but back-substitution is often employed when dealing with upper triangular systems, or when the system of equations has been previously transformed into a form where back-substitution becomes a straightforward approach.

In our exercise, after substituting the value of \(z\) into the second equation, we found the value of \(y\). Then we used both known values, \(y\) and \(z\), to solve for \(x\) in the first equation. This method is efficient whenever equations are already arranged in a way that allows you to 'peel back' the variables layer by layer, from the simplest equation to the more complex ones.

By the time we reach the final equation, it usually only contains one variable, simplifying the matter to solving a basic algebraic equation. After solving for each variable, it is essential to interpret the results correctly, as it confirms that the solution is valid for every equation in the system. The result from such a solution process is a set of values, one for each variable, which constitutes the solution to the system of linear equations.