Question

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} x+2 y-z=-3 \\ 2 x-4 y+z=-7 \\ -2 x+2 y-3 z=4 \end{array}\right. $$

Short answer

The solution is x = -3, y = 0.5, z = 1.

Step-by-step solution

Write the System of Equations

First, let's write down the system of equations we need to solve: \[ \begin{cases} x + 2y - z = -3 \ 2x - 4y + z = -7 \ -2x + 2y - 3z = 4 \end{cases} \]

Eliminate Variable z from Equations 1 and 2

To eliminate the variable z, add the first equation and the second equation:\[ (x + 2y - z) + (2x - 4y + z) = -3 + (-7) \]Simplifies to:\[ 3x - 2y = -10 \] We will call this Equation (4).

Eliminate Variable z from Equations 2 and 3

To eliminate the variable z from the second and third equations, add 3 times the second equation to the third equation:\[ 3(2x - 4y + z) + (-2x + 2y - 3z) = 3(-7) + 4 \]Simplifies to:\[ 4x - 10y = -17 \] We will call this Equation (5).

Solve System of Two Equations

Now solve the system formed by Equations (4) and (5): \[ \begin{cases} 3x - 2y = -10 \ 4x - 10y = -17 \end{cases} \]

Eliminate y

Multiply Equation (4) by 5 to align coefficients of y:\[ 5(3x - 2y) = 5(-10) \ 15x - 10y = -50 \] We will call this Equation (6).Now subtract Equation (5) from Equation (6):\[ (15x - 10y) - (4x - 10y) = -50 - (-17) \]\[ 11x = -33 \]\[ x = -3 \]

Find y

Substitute x = -3 into Equation (4):\[ 3(-3) - 2y = -10 \ -9 - 2y = -10 \ -2y = -1 \ y = \frac{1}{2} \]

Find z

Substitute x = -3 and y = \frac{1}{2} into the first equation:\[ (-3) + 2(\frac{1}{2}) - z = -3 \ -3 + 1 - z = -3 \ -2 - z = -3 \ -z = -1 \ z = 1 \]

Key concepts

solving systems of equations

Solving a system of equations means finding the values of the variables that satisfy all the given equations simultaneously. In this case, we have three equations involving three variables: x, y, and z. The ultimate goal is to determine the specific values of these variables that make all the equations true at the same time. Systems of equations can have:

• a single unique solution
• infinitely many solutions
• no solution (an inconsistent system)

We use different methods to solve these systems, such as the elimination method and the substitution method.

elimination method

The elimination method involves eliminating one variable at a time to simplify the system of equations. In the provided exercise, we focus on eliminating the variable z first. Let's see how this method works step-by-step:

• First, we add the first and second equations to eliminate z. This results in a new equation involving only x and y.
• Next, we eliminate z again by combining the second and third equations, giving us another equation with just x and y. Now, we have two equations with two variables.
• We then solve this new system of two equations to find the values of x and y.
• Finally, we substitute the found values of x and y back into one of the original equations to solve for z.

The elimination method requires careful algebraic manipulation to systematically reduce the complexity of the system until all variables are determined.

substitution method

The substitution method is another technique to solve systems of equations. It involves solving one equation for a variable and then substituting that expression into the other equations. Here’s a simplified outline of the substitution method:

• Solve one of the equations for one variable in terms of the others.
• Substitute this expression into the other equations to reduce the number of variables.
• Continue this process until you solve for one variable.
• Use the found value to back-substitute and find the other variables sequentially.

While not applied in the provided solution, the substitution method is particularly useful when one of the equations is easily solvable for one variable. It provides a straightforward path to hand calculations, especially in systems involving more complex or non-linear relations.

linear algebra

Linear algebra is a branch of mathematics focusing on linear equations and their representations through matrices and vector spaces. Understanding linear algebra concepts is crucial for solving systems of linear equations efficiently. Concepts such as matrix operations, determinants, and vector spaces are fundamental:

• Matrices provide a compact way to represent systems of linear equations.
• Determinants can help in finding the solution of a system or determining if a system has a unique solution.
• Vector spaces and linear transformations give insights into understanding the solution space of systems of equations.

In real-world applications, linear algebra techniques, often implemented through computational tools, offer powerful methods for solving large and complex systems of equations efficiently.