§ 3.3 Systems of Linear Equations in Three Variables

§ 3.3

Systems of Linear Equations in Three Variables

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.3

Systems of Equations in 3 Variables

In general, any equation of the form Ax + By + Cz = D where A, B, C, and D are real numbers such that A, B, C, and D are not all 0, is a linear equation in three variables x, y, and z.

The graph of this linear equation in three variables in a plane in three dimensional space.

A solution of a system of linear equations in three variables is an ordered triple of real numbers that satisfies all equations in the system.



Solving Linear Systems in 3 Variables by Eliminating Variables

1) Reduce the system to two equations in two variables. This is usually accomplished by taking two different pairs of equations and using the addition method to eliminate the same variable from both pairs.

2) Solve the resulting system of two equations in two variables using addition or substitution. The result is an equation in one variable that gives the value of that variable.

3) Back-substitute the value of the variable found in step 2 into either of the equations in two variables to find the value of the second variable.

4) Use the values of the two variables from steps 2 and 3 to find the value of the third variable by back-substituting into one of the original equations.

5) Check the proposed solution in each of the original equations.



EXAMPLEEXAMPLE

Show that the ordered triple (2,-1,3) is a solution of the system:

First Equation:

SOLUTIONSOLUTION

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

2x - y - 2z = -1

Substitute the given values

x + y + z = 4

2 + (-1) + 3 = 4

4 = 4 Simplifytrue


Systems of Equations in Application

CONTINUECONTINUEDD Second Equation:

Substitute the given values2 - 2(-1) - 3 = 1

1 = 1 Simplify

x - 2y - z = 1

Substitute the given values2(2) - (-1) – 2(3) = -1

1 = 1 Simplify

Third Equation:

2x - y - 2z = -1

Therefore, the ordered triple (2,-1,3) satisfies all three equations and is therefore a solution for the system of equations.

true

true



EXAMPLEEXAMPLE

Solve the system:

SOLUTIONSOLUTION

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

2x - y - 2z = -1

Equation 1Equation 2Equation 3

1) Reduce the system to two equations in two variables. There are multiple ways that we could proceed. We will use the addition method on equation’s 1 and 2 to eliminate x. Then we’ll eliminate the same variable, x, using equation’s 1 and 3.



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

Equation 3 No change

Multiply by -2 -2x - 2y - 2z = -8

Add: -3y - 4z = -9

CONTINUECONTINUEDD

2x - y - 2z = -1

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

Equation 1Equation 2

No change

Multiply by -1x + y + z = 4

-x + 2y + z = -1

Add: 3y + 2z = 3



CONTINUECONTINUEDD

Now I have a system of equations in two variables that I can solve.

3y + 2z = 3-3y - 4z = -9

Add: - 2z = -6z = 3 Divide both sides by -2

Now we can use one of the two equations in two variables to solve for y.

2) Solve the resulting system of two equations in two variables.



CONTINUECONTINUEDD

3y + 2z = 3

3y + 2(3) = 3

3y + 6 = 3

3y = -3y = -1

An equation in two variables

Replace z with 3

Multiply

Subtract 6 from both sides

Divide both sides by 3

3) Use back-substitution in one of the equations in two variables to find the value of the second variable.



CONTINUECONTINUEDD

4) Back-substitute the values found for two variables into one of the original equations to find the value for the third variable.

Now we can use any of the original equations to find x. Let’s use equation 2.

An equation in three variables

Replace z with 3 and y with -1

Multiply

Simplify

Add 1 to both sides

x - 2y - z = 1

x – 2(-1) – (3) = 1

x + 2 – 3 = 1

x -1 = 1x = 2



CONTINUECONTINUEDD

5) Check.

The previous example shows this step in its entirety.

Therefore, the solution is (2,-1,3) and the solution set is

{(2,-1,3)}.


Systems of Equations

Types of Systems of EquationsConsistent A system that has at least one solution

Inconsistent A system that is not consistent

(has no solutions)

Dependent A system that has infinitely many solutions

Independent A system that is not dependent

(has one or no solutions)



EXAMPLEEXAMPLE

The following is known about three numbers: Three times the first number plus the second number plus twice the third number is 5. If 3 times the second number is subtracted from the sum of the first number and 3 times the third number, the result is 2. If the third number is subtracted from the sum of 2 times the first number and 3 times the second number, the result is 1. Find the numbers.

SOLUTIONSOLUTION

First, we need to establish our variables.

Let x = the “first number”.Let y = the “second number”.Let z = the “third number”.



CONTINUECONTINUEDD

Three times the first number plus the second number plus twice the third number is 5.

Now we need to set up the equations. The first sentence says:

The corresponding equation is: 3x + y + 2z = 5The second sentence says:

The corresponding equation is: (x + 3z) – 3y = 2

If 3 times the second number is subtracted from the sum of the first number and 3 times the third number, the result is 2.

x + 3z – 3y = 2x - 3y + 3z = 2



CONTINUECONTINUEDD

1) Reduce the system to two equations in two variables. There are multiple ways that we could proceed. We will use the addition method on equation’s 1 and 2 to eliminate x. Then we’ll eliminate the same variable, x, using equation’s 2 and 3.

The third sentence says:

The corresponding equation is: (2x + 3y) – z = 1

2x + 3y - z = 1

If the third number is subtracted from the sum of 2 times the first number and 3 times the second number, the result is 1.



CONTINUECONTINUEDD

Equation 2Equation 3 No change

Multiply by -2

Add: 9y - 7z = -3

Equation 1Equation 2

No change

Multiply by -3

Add: 10y - 7z = -1

Equation 1Equation 2Equation 3

3x + y + 2z = 5x - 3y + 3z = 22x + 3y - z = 1

3x + y +2z = 5x - 3y + 3z = 2

x - 3y + 3z = 22x + 3y - z = 1

3x + y + 2z = 5-3x + 9y - 9z = -6

-2x + 6y - 6z = -42x + 3y - z = 1



CONTINUECONTINUEDD

Now I have a system of equations in two variables that I will solve.

Add: -y = -2

Now I can use one of the two equations in two variables to solve for z.

No Change

Multiply by -1 -10y + 7z = 1

y = 2

2) Solve the resulting system of two equations in two variables.

9y - 7z = -310y - 7z = -1

9y - 7z = -3



CONTINUECONTINUEDD

An equation in two variables

Replace y with 2

Multiply

Subtract 20 from both sides

Divide both sides by -7

3) Use back-substitution in one of the equations in two variables to find the value of the second variable.

10y - 7z = -1

10(2) - 7z = -1

20 - 7z = -1

- 7z = -21

z = 3



CONTINUECONTINUEDD

4) Back-substitute the values found for two variables into one of the original equations to find the value for the third variable.

Now I can use any of the original equations to find c. I’ll use equation 2.

An equation in three variables

Replace y with 2 and z with 3

Multiply

AddSubtract 3 from both sides

x - 3y + 3z = 2

x – 3(2) + 3(3) = 2

x – 6 + 9 = 2

x + 3 = 2x = -1



CONTINUECONTINUEDD

5) Check.

To check, we will plug the point into each equation.

(-1,2,3)

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Therefore, the solution is (-1,2,3).

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Therefore, the potential solution is

3x + y + 2z = 5 x - 3y + 3z = 2 2x + 3y - z = 13(-1) + (2) + 2(3) = 5 (-1) – 3(2) + 3(3) = 2 2(-1) + 3(2) – (3) = 1

-3 + 2 + 6 = 5 -1 – 6 + 9 = 2 -2 + 6 – 3 = 15 = 5 2 = 2 1 = 1

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§ 3.3 Systems of Linear Equations in Three Variables