Question

Solve each system by elimination. Check your answers

x + y + z = -1
2x - y + 2z = -5
-x + 2y - z = 4

Answer 1

The solution to the system of equations by elimination is found to be x = -2, y = 1, and z = 1. Each step is performed by systematically eliminating one variable at a time and back-substituting to find the remaining variables.

Step-by-step explanation:

The student has provided a system of three linear equations to be solved by elimination, which is a standard method used in algebra. In this method, we aim to eliminate one variable at a time by combining equations, which simplifies the system to two equations with two variables, and then finally one equation with one variable. We then solve for the remaining variables.

Firstly, we write down the given equations:

1. x + y + z = -1
2. 2x - y + 2z = -5
3. -x + 2y - z = 4

To eliminate one variable, we can add the first and third equations to eliminate x, yielding:

1. (x - x) + (y + 2y) + (z - z) = -1 + 4
2. 3y = 3
3. y = 1

Now substitute y = 1 into the first and second equations and eliminate z by adding the first and second modified equations to get the value of x:

1. x + 1 - z = -1 → x - z = -2 (i)
2. 2x - 1 + 2z = -5 → 2x + 2z = -4 → x + z = -2 (ii)

Adding equations (i) and (ii) we eliminate z:

1. (x + x) + (-z + z) = -2 + (-2)
2. 2x = -4
3. x = -2

Now we have x = -2 and y = 1, we can substitute these into any of the original equations to find z. Using the first equation:

1. -2 + 1 + z = -1
2. z = -1 + 2
3. z = 1

Therefore, the solution to the system by elimination is x = -2, y = 1, and z = 1. To check the solution, we can substitute these values back into the original equations to see if they hold true.