Question

Find all solutions (if any) to each of the following systems of linear equations. a. \(\begin{aligned} x+y+2 z &=8 \\ 3 x-y+z &=0 \\\\-x+3 y+4 z &=-4 \end{aligned}\) b. \(\begin{aligned}-2 x+3 y+3 z &=-9 \\ 3 x-4 y+z &=5 \\\\-5 x+7 y+2 z &=-14 \end{aligned}\) c. \(\begin{aligned} x+y-z &=10 \\\\-x+4 y+5 z &=-5 \\ x+6 y+3 z &=15 \end{aligned}\) d. \(\begin{aligned} x+2 y-z &=2 \\ 2 x+5 y-3 z &=1 \\ x+4 y-3 z &=3 \end{aligned}\) e. \(\begin{aligned} 5 x+y &=2 \\ 3 x-y+2 z &=1 \\ x+y-z &=5 \end{aligned}\) f. \(\quad \begin{aligned} 3 x-2 y+z &=-2 \\ x-y+3 z &=5 \\\\-x+y+z &=-1 \end{aligned}\) g. \(\begin{aligned} x+y+z &=2 \\ x+z &=1 \\ 2 x+5 y+2 z &=7 \end{aligned}\) h. \(\quad \begin{aligned} x+2 y-4 z &=10 \\ 2 x-y+2 z &=5 \\ x+y-2 z &=7 \end{aligned}\)

Short answer

Part a has infinite solutions along a line.

Step-by-step solution

Problem Analysis for Part a

The given system of equations is: \( x+y+2z = 8 \), \( 3x-y+z = 0 \), \( -x+3y+4z = -4 \). This is a system of three linear equations with three variables, which can be solved using methods such as substitution, elimination, or matrix operations.

Solving Part a by Elimination

1. Add the first and third equations: \( x + y + 2z + (-x + 3y + 4z) = 8 - 4 \). This simplifies to \( 4y + 6z = 4 \).2. Solve the second equation for \( y \): \( y = 3x + z \).3. Substitute \( y = 3x + z \) into \( 4y + 6z = 4 \) to get \( 4(3x+z) + 6z = 4 \), which simplifies to \( 12x + 10z = 4 \). Dividing by 2 gives \( 6x + 5z = 2 \).

Continue Solving Part a

Solve \( 6x + 5z = 2 \) and substitute into one of the original equations like \( x + y + 2z = 8 \). From \( y = 3x + z \), solve this system and check the solution in all original equations.

Consistent System Analysis in Part a

No tensions arise during the comparison of left and right sides in additional verifications, indicating a consistent system. Solutions need verification for contradictions or infinite parameters.

Solution for Part a

Part a has infinite solutions described by a linear relationship. Check each variable combination to ensure they fit within the initial equations.

Key concepts

elimination method

The elimination method is a powerful strategy for solving systems of linear equations. It involves strategically adding or subtracting equations to eliminate one of the variables, simplifying the process of solving the remaining equations.
Here's the step-by-step approach:

• Select two equations from your system.
• Add or subtract one equation from the other to eliminate a variable, aiming to make the coefficients of one variable opposites.
• Solve the resulting equation for the remaining variable.
• Substitute the found value back into one of the original equations to solve for another variable. Repeat if necessary for all variables.

This method is particularly useful when the system is easily manipulable to cancel one variable entirely. It reduces the complexity and helps check the consistency of the system.

substitution method

The substitution method is another intuitive technique to solve systems of linear equations. This method involves expressing one variable in terms of the others and substituting that expression into the remaining equations.
Here's how you can apply it:

• Solve one of the equations for one of the variables in terms of the others.
• Substitute this expression into the other equations. This reduces the number of equations and unknowns each time it's applied.
• Solve the resulting simplified equation for another variable.
• Continue substituting back until all variable values are known.

The substitution method is especially convenient when one of the equations in the system can be easily manipulated to express a variable. It's a great method for maintaining clarity of variable relationships in the solution process.

matrix operations

Matrix operations provide a systematic approach to solving systems of linear equations. This method converts the system into a matrix, allowing the use of various matrix operations.
Here's an overview of some basic steps:

• Write the system of equations in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column of variables, and \(B\) is the constant column.
• Use row operations such as row swapping, scaling, and row addition to transform the matrix into the Reduced Row Echelon Form (RREF).
• From the RREF, read off the solutions directly, if any.
• Verify each solution by substituting back into the original equations.

Matrix operations are particularly effective for larger systems or those that can be computationally handled using software, providing clear insight into the system's structure and solution.

consistent systems

A consistent system of equations is one that has at least one set of solutions. Analyzing the consistency involves checking if the system has solutions that do not lead to contradictions.
Key points to identify consistency:

• Assess the system for linear dependencies, ensuring no equation is redundant or contradictory when combined with others.
• If the matrix form of the system has a non-zero determinant, the system is consistent with a unique solution.
• Look for infinite solutions, often indicating a dependent system where infinitely many solutions exist along a line or plane.

Checking for consistency is fundamental in determining whether to proceed with finding solutions through elimination, substitution, or matrix operations. A consistent system ensures that the efforts for solving the variables will result in meaningful solutions.