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In mathematical terms, a system of equations comprises multiple equations involving the same set of variables. The main goal when dealing with a system is to find a common solution, which satisfies all equations in the system. In simpler terms, we are looking for a set of values that makes each equation true.
Different methods can be used to solve these systems, including substitution, elimination, and graphing.
For example, in the given system:

• \(2u - 3v = 5\)
• \(2v = 6\)

we can use substitution because one equation is already easily solvable for one variable. The process of solving a system often leads us to utilize such methods, based on the structure and simplicity of the equations involved.

Linear algebra is an essential area of mathematics focused on vector spaces and linear mappings between them. Specifically, it deals with lines, planes, and subspaces, and is used extensively in fields like engineering, physics, computer science, and economics.

An essential application of linear algebra is solving systems of linear equations, such as the one provided in the original exercise. In this system, the emphasis on linear expressions (coefficients and constants) simplifies the process of ascertain the relationship between variables like \(u\) and \(v\).
Key concepts in linear algebra include:

• Vectors and matrices
• Determinants and eigenvalues
• Vector spaces and linear transformations

Understanding these concepts can greatly aid in grasping how solutions to systems are developed. In this case, the system of equations can be thought of in terms of matrix operations, though the given system was straightforwardly solved using substitution.

Back substitution is a method used primarily when solving systems of linear equations by substitution or elimination methods. It is especially useful in scenarios involving upper triangular matrices as seen in Gaussian elimination.

In the original exercise, back substitution is clearly shown through step-by-step solving:

• First, solve one of the equations for a variable, making it simplest to isolate one variable. Here, by solving \(2v = 6\), we find \(v = 3\).
• This solved value is then substituted back into the other equation, which has the unknown variable \(u\), resulting in an equation with one variable: \(2u - 9 = 5\).
• Solve this new equation by isolating \(u\), providing the final answer, \(u = 7\).

This straightforward approach illustrates how back substitution can systematically unravel complex equations.