Problem 16 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use back-substitution to solve the system of linear equations. $$\left\\{\begin{aligned} x-y+2 z &=22 \\ 3 y-8 z &=-9 \\ z &=-3 \end{aligned}\right.$$

Step-by-Step Solution

Answer

After using back-substitution, the solutions to the system of equations are: x = 14, y = -11, z = -3.

Step 1: Identify the given value

In the given system of linear equations, 'z' is already given as -3.

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

The second equation is 3y - 8z = -9. Substituting z = -3, we get 3y - 8*(-3) = -9, which simplifies to 3y = -9 -24, yielding y = -11.

Step 3: Substitute the values of 'y' and 'z' into the first equation

The first equation is x - y + 2z = 22. Substituting y = -11 and z = -3, we get x - (-11) + 2*(-3) = 22, which simplifies to x = 22 - 8, hence x = 14.

Key Concepts

Back-Substitution Method

The back-substitution method is a straightforward technique used to solve systems of linear equations where at least one variable is already known or can be easily determined. In our exercise, the system of equations is organized in a way that allows us to start from the last equation, which provides the value of the variable 'z' directly, being 'z = -3'.

With this value, we move upwards through the other equations and progressively substitute the known values to find the remaining unknowns. It's akin to unraveling a thread: once you have a starting point, the rest follows naturally step by step.
It's especially efficient when the system is in upper triangular form, where each equation above the last provides one less unknown, as seen in the exercise. This is why it's called 'back'-substitution - we start from the 'back' of the system (the last equation) and work our way to the 'front' (the first equation).

This method is generally taught in linear algebra courses and is a fundamental skill for anyone studying mathematics, physics, engineering, and other sciences where such systems often arise.

Linear Algebra

Linear algebra is an area of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It is fundamental in various fields such as engineering, physics, computer science, and economics because many real-world problems can be modeled using linear structures.

In the context of our exercise, linear algebra provides the theoretical backbone for methods like the back-substitution method. It teaches how systems of equations can represent vector equations or matrix equations. Understanding the principles of linear algebra can help you visualize these problems not just as numbers and variables, but as shapes, directions, and dimensions in space.

Linear algebra gives us the tools, such as matrices and determinants, to solve these systems more methodically and understand the underlying relationships between the variables. It's truly the language of organized numerical relationships and spatial concepts.

Substitution in Equations

Substitution in equations is a technique where we replace one variable with its equivalent value, often derived from another equation in a system. It's a fundamental concept in algebra that enables the solution of numerous complex problems.

In the exercise, once 'z' was identified to be -3 from the third equation, we then 'substituted' this value into the second equation to find 'y'. Similarly, with the values of 'z' and 'y' known, we substituted both into the first equation to find 'x'.

This process underscores the interconnectedness of variables within systems and the importance of isolating and solving for one variable at a time. In some cases, substitution might require algebraic manipulation like distributing, combining like terms, or isolating variables, to pave the way for replacing the variables. It is a vital skill that provides a foundation for solving not only simple linear equations but complex equations across various fields of mathematics and science.