Solving a system of three equations in three unknowns is more complex than solving a system of two equations, but it uses similar concepts. In solving a system of three equations, we'll combine equations in pairs to reduce them to equations in two variables, and then we'll use those two equations to solve for two of the variables, and finally we'll plug those values back into one of the original equations in order to find the remaining variable. Sounds confusing? Let's try an example!

Example

Equation 1: x + y + z = 6
Equation 2: x  - y + 2z = 5
Equation 3: 2x + 3y - z = 5

Solution

I've numbered the equations 1, 2, and 3, because it will help us to keep track of what we've done. First, we'll look for two equations that we can combine to easily get rid of a variable. I notice that if we add equations 1 and 2, the y values will cancel out:

Combining equations 1 and 2: If we simply add the equations, we get 2x + 3z = 11

Now we look for another pair of equations that we can use to eliminate the same variable. I notice that if I use equations 2 and 3, I can multiply the second equation by 3, and then add them:

3x - 3y + 6z = 15
2x + 3y -  z =  5
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5x + 5z      = 20
x + z = 4

Solve for x and z using the system of equations 2x + 3z = 11 and x + z = 4:

x = 4 - z (solving the second equation for x)
2(4 - z) + 3z = 11 (substituting into the first equation)
8 - 2z + 3z = 11
z = 3

From this, we know x = 4 - z = 4 - 3 = 1

Knowing that x = 1 and z = 3, we can plug these values into any of the original equations in order to solve for y.

x + y + z = 6
1 + y + 3 = 6
y + 4 = 6
y = 2

Thus, the solution is: x = 1; y =2; z = 3

1.

Describe, in your own words, the process for solving a system of three equations in three unknowns

2.

Solve the system: x + y + z = 5; x - y - z = -3; x + 2y + z = 6

3.

Solve the system: x + y + z = 2; x - y - z = 4; 2x + 3y - z = -5

4.

Solve the system: x + 2y + 3z = 7; 2x +3y - z = 11; 3x - 2y + z = -3

5.

Solve the system: x + y = 5; y + z = 8; x + z = 7