Solving Systems of Equations with Three Variables

Solving Systems of Equations with Three Variables

Recall the following about solving systems of equations:

- A
**consistent system**is a system that has**at least one solution.** - An
**inconsistent system**is a system that**has no solution**.

The equations of a system are **dependent** if ALL the solutions of one equation are also solutions of the other two equations. In other words, they end up being **the same equation**.

The equations of a system are **independent** if they **do not share ALL solutions**. They can have multiple solutions on a separate equation, but all three of them don't coincide.

When solving systems of equations in three variables, it is generally recommended to use the elimination method to eliminate a variable. This will need to be repeated until all variables are determined.

**Example**

Solve the following system:

x + y + z = 4

x - 2y - z = 1

2x - y - 2z = -1

First, let's eliminate x. This can be done by taking the equations two at a time and using the elimination method to cancel out x each time.

Start with the first and second equations. Multiply the first equation by -1 and add it to the second equation to eliminate x.

-x - y - z = -4

x - 2y - z = 1

-3y - 2z = -3

Now let's use the first and third equations. Multiply the first equation by -2 and add it to the third equation.

-2x - 2y - 2z = -8

2x - y - 2z = -1

-3y - 4z = -9

Next, use the two equations created by those steps to eliminate y. Multiply the first equation by -1 and add it to the second.

3y + 2z = 3

-3y - 4z = -9

-2z = -6

Solve the resulting equation for z.

-2z = -6

z = 3

Substitute 3 for z in one of the two y-z equations and solve for y.

-3y - 2z = -3

-3y - 2(3) = -3

-3y - 6 = -3

-3y = 3

y = -1

Finally, substitute -1 for y and 3 for z in one of the original equations and solve for x.

x + y + z = 4

x + (-1) + 3 = 4

x + 2 = 4

x = 2

The answer is (2, -1, 3).