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The substitution method is a way to solve a system of equations by expressing one variable in terms of the others and then substituting that expression into the other equations. In the given system, we could choose one of the equations and solve it for a single variable. For example, we might solve the first equation for \( x \) as follows:\[x = -1 - 2y + z\]This expression for \( x \) could then be substituted into the other two equations in the system. The goal is to reduce the number of variables in those equations, eventually simplifying to a system where each equation contains only two variables. This method can be more straightforward when one of the equations is easier to solve for one of the variables. However, it often involves more calculations if the algebra becomes complex when substituting expressions into the other equations.

The elimination method involves adding or subtracting equations to eliminate a common variable, simplifying the system to fewer equations with fewer variables. For our exercise, elimination is used to remove the variable \( x \) from the first two equations by combining them efficiently:

• Multiply the first equation by a factor to align the \( x \)-coefficients with the second equation.
• Subtract or add these equations to cancel out \( x \).

For our given exercise:- Multiply the first equation by 3: \( 3(x + 2y - z) = 3(-1) \rightarrow 3x + 6y - 3z = -3 \)- Subtract this new equation from the second one: \( (3x - 4y + 5z) - (3x + 6y - 3z) = 9 - (-3) \rightarrow -10y + 8z = 12 \)This results in a simpler system with fewer variables, making it easier to solve.

Matrix algebra can be an efficient tool for solving systems of linear equations. This approach involves representing the system in matrix form and using techniques such as row reduction or matrix inverses. The system can be written in the form:\[AX = B\]Where \( A \) is the matrix of coefficients, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants. Using the matrix for our system:\[A = \begin{bmatrix} 1 & 2 & -1 \ 3 & -4 & 5 \ -5 & 1 & -7 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} -1 \ 9 \ -21 \end{bmatrix}\]Matrix methods, such as Gaussian elimination or the use of inverse matrices, can then be applied to find the values of \( x \), \( y \), and \( z \). While matrix algebra is powerful for larger systems, it may be more complex than simple substitution or elimination for smaller systems.

Solving a system of equations involves a series of methodical steps to find the values of the variables. Let's encapsulate these steps:

• Identify and confirm the system of equations, ensuring they are linear and appropriate for the method chosen.
• Choose the most suitable method: either substitution, elimination, or matrix algebra based on the problem context and complexity.
• Simplify the system by eliminating variables, either by substitution or adding/subtracting equations in the elimination method.
• Solve the resulting simpler system of equations to find the values of the remaining variables.
• Back-substitute these values into the original equations to find the values of all variables.
• Verify the solution by plugging the variable values back into the original equations to ensure all equations are satisfied.

Each step is crucial to ensure a correct and complete solution, emphasizing understanding over rote computation.