Question

Solve the system using the elimination method. \(x+2 y-z=3\) \(-2 x-y+z=-1\) \(6 x-3 y-z=-7\)

Short answer

The solution to the system of equations is \(x = 0.5\), \(y = 2.5\), and \(z = 1\).

Step-by-step solution

Add the First and Second Equations

To eliminate z, add the first and second equations together: \((x+2y-z) + (-2x-y+z) = 3+(-1)\) to get \(-x + y = 2\)

Multiply and Add the Third and First Equations

Again, to eliminate z, multiply the first equation by 3 and add it to the third equation: \(3(x+2y-z) + 6x -3y-z = 3*3 + (-7)\) to get \(9x-y = 2\)

Solve the System of Two Equations

The system of two equations from steps 1 and 2 is \(-x + y = 2\) and \(9x - y = 2\). Adding those two equations together, results, \(8x=4 \Rightarrow x = 0.5\) . Substitute \(x = 0.5\) into the first equation from step 1 to solve for y: \(-0.5 + y = 2 \Rightarrow y = 2.5\)

Substitute to Find z

Substitute \(x = 0.5\) and \(y = 2.5\) into the first original equation to solve for z: \(0.5 + 2 * 2.5 - z = 3 \Rightarrow z = 1\)

Key concepts

Systems of Equations

A system of equations is a set of two or more equations with the same variables. The solutions to a system of equations are the value(s) that satisfy all equations simultaneously. In our case, we have a system of three equations with three unknowns (x, y, and z). Solving such systems can be complex, but there are several methods—such as graphing, substitution, and elimination—that can simplify the process. The elimination method, which is used in our exercise, involves combining equations in a way that cancels out one or more variables, allowing for easier solution of the reduced system.

Algebraic Techniques

Algebraic techniques are methods or procedures used to manipulate algebraic expressions and solve equations or systems of equations. These techniques include operations like addition, subtraction, multiplication, division, and the application of the distributive property. To effectively use these techniques, it's crucial to understand the properties of equality and the rules for operating with variables.

For our exercise, the elimination method called upon a few algebraic maneuvers. First, we combined equations to eliminate the variable z. This involved simple addition of the equations as they stood or multiplication of the entire equation by a constant to facilitate the elimination of a variable. In the next step, we used addition again to eliminate another variable, eventually reducing the system to a single variable equation which could be easily solved. Understanding how to manipulate the equations without changing their fundamental properties is essential in systematically reducing the complexity of the original system.

Substitution Method

While the substitution method wasn't used to solve this particular exercise, it's another effective tool for solving systems of equations. In substitution, you solve one equation for a variable and then 'substitute' this expression into the other equation(s). This method works well for systems where one equation is easier to solve for one variable than the others, or in systems where substituting directly simplifies the system quickly. However, this method can become cumbersome if the substitution does not lead to simpler equations.

Still, it's important to note that the substitution method can sometimes be used in conjunction with the elimination method, as seen in the final steps of our example problem. After using elimination to reduce the system to two equations, we 'substituted' the value of x found into one of the earlier equations to find the value of y, and eventually z. Mastering these methods provides a strong foundation for tackling more complex algebraic challenges.