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An algebraic solution involves using mathematical operations to find the values of unknown variables in an equation or set of equations. When solving systems of linear equations, the goal is to find values for variables that satisfy all equations in the system simultaneously.
In algebra, methods like substitution and elimination can be utilized, which involve logical steps to gradually reduce the number of variables, making the equations simpler and more manageable. These techniques rely on basic arithmetic operations like addition, subtraction, multiplication, and division.
This approach is particularly powerful for linear systems, where equations represent lines on a graph, and the solution is the point(s) where these lines intersect. Solving such systems algebraically avoids the need for graphing, offering a precise and clear method to find solutions.

The elimination method is a popular technique for solving systems of linear equations. It involves removing one of the variables to make the system simpler. This is usually done by adding or subtracting equations from one another.

• Identify a variable to eliminate.
• Multiply each equation if necessary to align coefficients of the chosen variable.
• Subtract or add equations to eliminate the variable, creating a new equation with one less variable.

In our example, subtracting the second equation from the first eliminated the variable \(y\). This simplification produced a new equation with only \(x\) and \(z\), which could then be solved more easily.

Variable substitution is another effective approach to solve systems of equations. This technique involves expressing one variable in terms of the others and substituting that expression into other equations.

• Start with a chosen equation and solve it for one variable.
• Replace the chosen variable in the other equations with the entity found in the first step.
• This results in equations that have fewer variables, making them easier to solve.

In our solution, the third equation provided an expression for \(y\) in terms of \(z\), \(y = (13 - 4z) / 3\). By substituting this back into the remaining equations, the problem was reduced to solving for two variables: \(x\) and \(z\). This method simplifies the system, leading to the discovery of the specific values for each variable that satisfies the entire system.