91Ó°ÊÓ

A system of linear equations consists of multiple equations that share common variables. The core objective is to find the set of values that satisfy all equations simultaneously. Such a system can be represented in matrix form, but is often presented as aligned equations, like the one in our exercise:
\[\begin{aligned} 4x - 2y + z &= 8 \ -y + z &= 4 \ z &= 11 \end{aligned}\]
These equations graphically represent lines in two dimensions or planes in three dimensions. The solution to this system is where all lines or planes intersect. Depending on context, a system can have exactly one solution (intersecting at a point), no solution (parallel, never intersecting), or infinitely many solutions (overlapping). The system provided in the exercise is designed to have a unique solution, which we find using back-substitution.

Solving linear equations is a foundational skill in algebra. The process involves manipulating the equation by applying arithmetic operations in an effort to isolate the variable on one side of the equation. We employ several strategies to tackle these equations:

• Combining like terms
• Using inverse operations
• Distributive property

For example, if we have an equation like \(2x + 3 = 7\), we would subtract 3 from both sides to get \(2x = 4\), and then divide by 2 to find \(x = 2\). In our textbook problem, we solve progressively by back-substitution. We start with the last equation to find the value of one variable, and then use this value to solve for the others progressively.

The substitution method is a technique to solve a system of linear equations, where we first solve one equation for one variable and then substitute this solution into the other equations. This method is particularly efficient when one of the equations is already solved for a variable or can be easily manipulated to do so.
Here are steps we follow in substitution:

Isolate a Variable

In the given exercise, the equation \(z = 11\) is already isolated. We did not need to perform any operations to solve for \(z\).

Substitute and Solve

Next, we substitute \(z = 11\) into the other equations and solve for the remaining variables, one at a time. In our case, we plugged it into the second equation to find \(y\), and then into the first to solve for \(x\).
This systematic approach minimizes errors and efficiently breaks down a complex problem into simpler steps, leading to the solution of the entire system.