Identifying Solutions

A solution of a system of three equations in three variables is an ordered triple (x, y, z) that makes all three equations true.

A linear equation in three variables is an equation equivalent to one in the form Ax + By + Cz = D, where A, B, C, and D are real numbers. We refer to the form Ax + By + Cz = D as standard form for a linear equation in three variables.

Example Determine whether (2, –1, 3) is a solution of the system:

4

2 2 3

4 2 3

x y z

x y z

x y z

Solution We substitute (2, –1, 3) into the three equations, using alphabetical order:

x + y + z = 4

2 + (–1) + 3 4

4 = 4 3 = 3TRUE TRUE

The triple makes all three equations true, so it is a solution of the system.

2x – 2y – z = 3

2(2) – 2(–1) – 3 3

– 4x + y + 2z = –3

– 4(2) + (–1) + 2(3) –3

–3 = –3 TRUE

Solving Systems in Three Variables

The elimination method allows us to manipulate a system of three equations in three variables so that a simpler system of two equations in two variables is formed. Once that simpler system has been solved, we can substitute into one of the three original equations and solve for the third variable.

Solution

We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2): 6

2 2

x y z

x y z

(1)(2)(4)2x + 3y = 8 Adding to eliminate z

Example

Solve the following system of equations:

6

2 2

3 8

x y z

x y z

x y z

(1)(2)(3)

Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z.

2 2

3 8

x y z

x y z

(5)

x – y + 3z = 8

Multiplying equation (2) by 3

4x + 5y = 14

3x + 6y – 3z = 6

Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system,

4x + 5y = 14

2x + 3y = 8

(5)

(4)

We multiply both sides of equation (4) by –2 and then add to equation (5):

Substituting y = 2 into either equation (4) or (5) we get,

4x + 5y = 14

–4x – 6y = –16

–y = –2

y = 2

Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.

x = 1

4x + 5(2) = 14

4x + 10 = 14

4x = 4

Let’s use equation (1) and substitute our two numbers in it:

We have obtained the ordered triple (1, 2, 3). It should check in all three equations.

x + y + z = 6

1 + 2 + z = 6

z = 3

The solution is (1, 2, 3).

Dependency, Inconsistency, and Geometric Considerations

The graph of a linear equation in three variables is a plane. Solutions are points common to the planes of each system. Since three planes can have an infinite number of points in common or no points at all in common, we need to generalize the concept of consistency in three dimensions.

Solution

The variable x is missing in equation (1). Let’s add equations (2) and (3) to get another equation with x missing:

2 5

1

x y z

x y z

(2)(3)(4)y + 2z = 6 Adding

Example

Solve the following system of equations:

2 2

2 5

1

y z

x y z

x y z

(1)(2)(3)

Equations (1) and (4) form a system in y and z. Solve as before:

Multiplying equation (1) by –1

0 = 4

y + 2z = 6

y + 2z = 2 –y – 2z = –2

y + 2z = 6

Since we ended up with a false equation, or contradiction, we know that the system has no solution.

The system is inconsistent.

Examples: Solve each system

4 4

2 1

6 3 2 3

x y z

x y z

x y z

3 6 2

2 2 3 3 0.5

1

r s t

r s t

r s t

7

3 2 9

4 5

l m

m n

l n