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A system of equations is a collection of two or more equations with the same set of unknowns. In this example, we are given a simple system of two linear equations.
Each equation in the system provides information about the relationship between the variables.

• The first equation tells us how the variables are related: \(x - 2y = 1\).
• The second equation provides the value of one of the variables directly: \(y = 3\).

The goal of solving a system of equations is to find values for the variables that make all the equations true at the same time. Here, the task is simplified because one of the equations provides us with a specific value for \(y\).
To solve systems of equations, we typically look for ways to eliminate one variable in order to solve for another. This can be done through techniques like substitution or elimination.

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear equations. In the context of solving systems of equations, linear algebra provides the tools to systematically approach finding solutions.
A linear equation, like \(x - 2y = 1\), represents a straight line when graphed on a coordinate plane. When solving a system of such equations, we seek the intersection points of these lines.

• If the lines intersect at a single point, it indicates a unique solution, as in our example.
• If the lines are parallel, there might be no solution.
• If the lines coincide, there are infinitely many solutions.

Understanding how to work with linear equations is a critical component of linear algebra, and back substitution is one method to find solutions especially when the system is reducing to simpler forms.
In linear algebra, solving a system by back substitution commonly involves starting with the simplest equation and using it to find the values of unknowns in more complex equations.

Solving a system of equations using back substitution involves a series of deliberate steps. This method is particularly effective when one of the variables is already isolated in one of the equations, as seen here with \(y = 3\). To solve the system:

• First, we identify known values. Here, we know \(y = 3\) has already been solved for us.
• Next, we substitute this known value into the other equation. So, substitute \(y = 3\) into the equation \(x - 2y = 1\) to get \(x - 2(3) = 1\).
• Simplify the equation by performing arithmetic operations. In this case, multiply \(2\) by \(3\) to get \(6\), leading to \(x - 6 = 1\).
• Finally, solve for the remaining variable by isolating it. Add \(6\) to both sides to find \(x = 7\).

Each of these steps methodically brings us closer to the solution. By executing these actions one by one, complex systems of equations become manageable and easier to solve.