Problem 12 - Precalculus with Limits

Question

Using Back-Substitution In Exercises \(11 - 16 ,\) use back-substitution to solve the system of linear equations. $$\left\\{ \begin{aligned} 4 x - 3 y - 2 z & = 21 \\ 6 y - 5 z & = - 8 \\ z & = - 2 \end{aligned} \right.$$

Step-by-Step Solution

Answer

The solutions to the system of linear equations are \(x = 2\), \(y = -3\) and \(z = -2\).

Step 1: Evaluate equation 3

The third equation is \(z = -2\). This is the first solution as there are no other variables in this equation.

Step 2: Substitute z's value into equation 2

The second equation is \(6y - 5z = -8\). Substitute the value of \(z\) from equation 3 into this equation to get \(6y - (-10) = -8\), we simplify to get \(6y = -18\) and by dividing by 6 we get \(y = -3\).

Step 3: Substitute z's and y's value into equation 1

The first equation is \(4x - 3y - 2z = 21\). Substitute the values of \(z\) and \(y\) into this equation we get \(4x - (-9) - (-4) = 21\), simplifying we get \(4x = 8\) and on dividing by 4 we get \(x = 2\).

Key Concepts

Back-Substitution

Back-substitution is a powerful method for solving systems of linear equations, especially when the system is already in upper triangular form. This technique involves working from the last equation upwards to find the solution for each variable step by step. In the given exercise, the system of equations is:

• \( 4x - 3y - 2z = 21 \)
• \( 6y - 5z = -8 \)
• \( z = -2 \)

We start with the equation containing the fewest variables, usually the last equation. The third equation gives us the value of \(z\) directly, as \( z = -2 \).
Next, we substitute \(z = -2\) into the second equation to find \(y\). By solving \(6y - 5(-2) = -8\), we determine \(y = -3\).
Finally, having both \(y\) and \(z\) values, we use them in the first equation to solve for \(x\). Substituting \(y = -3\) and \(z = -2\) into the first equation, we solve \(4x - (-9) - (-4) = 21\), resulting in \(x = 2\).
Using back-substitution ensures that each variable is isolated and solved in the correct order, making it a step-by-step and logical approach.

Linear Equations

Linear equations are mathematical expressions that represent straight lines when graphed. They have the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are constants, and \(x\), \(y\), \(z\) are variables depicting the coordinates in space. In this exercise, we encounter three linear equations:

• \(4x - 3y - 2z = 21\)
• \(6y - 5z = -8\)
• \(z = -2\)

Each equation helps to describe the relationship between different variables in a three-dimensional space. By solving these equations simultaneously, we can determine a single set of values for \(x\), \(y\), and \(z\) that satisfy all equations at once.
The unique aspect of linear equations is their simplicity. They involve no variables raised to any power other than one, which gives them their characteristic straight-line representation. Their linear nature makes it feasible to use algebraic methods like back-substitution for finding solutions efficiently.

Algebraic Solutions

Finding algebraic solutions is essential for addressing systems of equations. This means using algebraic techniques to find explicit values for variable quantities within a set of equations. In this context, the core technique applied is substitution and simplification.
To begin, you solve for one variable in the simplest equation. For instance, \(z\) is solved immediately in \(z = -2\). This value is then used in other equations to simplify them one by one. For example, plugging \(z = -2\) into \(6y - 5z = -8\) allowed us to find \(y = -3\).
By continuing this method, substituting known values from one step into the next equation, you systematically find all unknowns. Lastly, substituting both \(y\) and \(z\) into the first equation, we determine \(x = 2\).
This process is effective because it builds step-by-step on known truths, using prior results to simplify each remaining equation. Such a structured approach ensures that every variable is resolved in an orderly fashion, ultimately arriving at the algebraic solutions effortlessly.