4.4 Solve Systems of Equations with Three Variables

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let's review what we already know about solving equations and systems involving up to two variables.

We learned earlier that the graph of a linear equation, $a x + b y = c,$ is a line. Each point on the line, an ordered pair $(x, y),$ is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions

We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Figure shows three graphs. In the first one, two lines intersect. Intersecting lines have one point in common. There is one solution to this system. The graph is labeled Consistent Independent. In the second graph, two lines are parallel. Parallel lines have no points in common. There is no solution to this system. The graph is labeled inconsistent. In the third graph, there is just one line. Both equations give the same line. Because we have just one line, there are infinitely many solutions. It is labeled consistent dependent.

Similarly, for a linear equation with three variables $a x + b y + c z = d,$ every solution to the equation is an ordered triple, $(x, y, z)$ , that makes the equation true.

All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, we’ll find the solution to the system.

When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.

Eight figures are shown. The first one shows three intersecting planes with one point in common. It is labeled Consistent system and Independent equations. The second figure has three parallel planes with no points in common. It is labeled Inconsistent system. In the third figure two planes are coincident and parallel to the third plane. The planes have no points in common. In the fourth figure, two planes are parallel and each intersects the third plane. The planes have no points in common. In the fifth figure, each plane intersects the other two, but all three share no points. The planes have no points in common. In the sixth figure, three planes intersect in one line. There is just one line, so there are infinitely many solutions. In the seventh figure, two planes are coincident and intersect the third plane in a line. There is just one line, so there are infinitely many solutions. In the last figure, three planes are coincident. There is just one plane, so there are infinitely many solutions.

Three parallel planes demonstrate an inconsistent system with no solution. As parallel planes never intersect, they have no points in common.

Text states 'Two planes are coincident and parallel to the third plane. The planes have no points in common.' An illustration shows two distinct parallel planes, highlighting the contradictory nature of the statements.

Two parallel planes (blue) are shown, each intersected by a third plane (orange). The illustration emphasizes that the two parallel planes themselves have no points in common.

Three planes are depicted, intersecting pairwise with each other. Notably, there is no single point where all three planes converge, illustrating a unique geometric relationship.

Three planes intersect along a single line, illustrating a consistent system with dependent equations and infinitely many solutions, as there are endless points along the shared line.

Depiction of two planes intersecting, illustrating a scenario where two coincident planes intersect a third in a line, resulting in infinitely many solutions as described by the text.

Three coincident planes are represented by a single blue plane, illustrating that there is only one distinct plane and thus infinitely many solutions.

To solve a system of three linear equations, we want to find the values of the variables that are solutions to all three equations. In other words, we are looking for the ordered triple $(x, y, z)$ that makes all three equations true. These are called the solutions of the system of three linear equations with three variables.

To determine if an ordered triple is a solution to a system of three equations, we substitute the values of the variables into each equation. If the ordered triple makes all three equations true, it is a solution to the system.

Solve a System of Linear Equations with Three Variables

To solve a system of linear equations with three variables, we basically use the same techniques we used with systems that had two variables. We start with two pairs of equations and in each pair we eliminate the same variable. This will then give us a system of equations with only two variables and then we know how to solve that system!

Next, we use the values of the two variables we just found to go back to the original equation and find the third variable. We write our answer as an ordered triple and then check our results.

The steps are summarized here.

How To

Solve a system of linear equations with three variables.

Write the equations in standard form
- If any coefficients are fractions, clear them.
Eliminate the same variable from two equations.
- Decide which variable you will eliminate.
- Work with a pair of equations to eliminate the chosen variable.
- Multiply one or both equations so that the coefficients of that variable are opposites.
- Add the equations resulting from Step 2 to eliminate one variable
Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
The two new equations form a system of two equations with two variables. Solve this system.
Use the values of the two variables found in Step 4 to find the third variable.
Write the solution as an ordered triple.
Check that the ordered triple is a solution to all three original equations.

When we solve a system and end up with no variables and a false statement, we know there are no solutions and that the system is inconsistent. The next example shows a system of equations that is inconsistent.

When we solve a system and end up with no variables but a true statement, we know there are infinitely many solutions. The system is consistent with dependent equations. Our solution will show how two of the variables depend on the third.

Example 5

Solve the system of equations: ${\begin{matrix} x + 2 y - z = 1 \\ 2 x + 7 y + 4 z = 11 \\ x + 3 y + z = 4 \end{matrix} .$

${\begin{matrix} x + 2 y - z = 1 (1) \\ 2 x + 7 y + 4 z = 11 (2) \\ x + 3 y + z = 4 (3) \end{matrix}$

Use equation (1) and (3) to eliminate x.

Use equation (1) and (2) to eliminate x again.

Use equation (4) and (5) to eliminate $y$ .

	There are infinitely many solutions.
Solve equation (4) for y.	Represent the solution showing how x and y are dependent on z. $\begin{matrix} y + 2 z & = & 3 \\ y & = & −2 z + 3 \end{matrix}$
Use equation (1) to solve for x.	$x + 2 y - z = 1$
Substitute $y = −2 z + 3 .$	$\begin{matrix} x + 2 (−2 z + 3) - z & = & 1 \\ x - 4 z + 6 - z & = & 1 \\ x - 5 z + 6 & = & 1 \\ x & = & 5 z - 5 \end{matrix}$

The true statement $0 equals 0$ tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form $(x, y, z)$ where $x = 5 z - 5; y = −2 z + 3$ and z is any real number.

Solve Applications using Systems of Linear Equations with Three Variables

Applications that are modeled by a systems of equations can be solved using the same techniques we used to solve the systems. Many of the application are just extensions to three variables of the types we have solved earlier.

Example 6

The community college theater department sold three kinds of tickets to its latest play production. The adult tickets sold for $15, the student tickets for $10 and the child tickets for $8. The theater department was thrilled to have sold 250 tickets and brought in $2,825 in one night. The number of student tickets sold is twice the number of adult tickets sold. How many of each type did the department sell?

We will use a chart to organize the information.

A table displays ticket pricing details for adults, students, and children. It shows the number of each type (x, y, z), their individual values ($15, $10, $8), and their total values (15x, 10y, 8z). The total number of tickets is 250, and the total value is $2825.

Number of students is twice number of adults.

Rewrite the equation in standard form.

\begin{matrix} y & = & 2 x \\ 2 x - y & = & 0 \end{matrix}

The image displays a mathematical problem with the prompt 'Write the system of equations.' To the right, a system of three linear equations with variables x, y, and z is presented.

Use equations (1) and (2) to eliminate z.

Solving a system of linear equations by multiplying the first equation by -8 and adding it to the second equation to eliminate 'z', resulting in 7x + 2y = 825.

Use (3) and (4) to eliminate

y .

A step-by-step solution demonstrating how to solve a system of linear equations by elimination. The first equation (-2x + y = 0) is multiplied by -2 to yield 4x - 2y = 0, which is then added to the second equation (7x + 2y = 825) to eliminate 'y', resulting in 11x = 825.

Solve for x.

x equals 75

adult tickets

Use equation (3) to find y.

−2 x + y = 0

Substitute

x equals 75.

\begin{matrix} −2 (75) + y & = & 0 \\ −150 + y & = & 0 \\ y & = & 150 student tickets \end{matrix}

Use equation (1) to find z.

x + y + z = 250

Substitute in the values

x = 75, y = 150 .

\begin{matrix} 75 + 150 + z & = & 250 \\ 225 + z & = & 250 \\ z & = & 25 child tickets \end{matrix}

Write the solution.

The theater department sold 75 adult tickets,
150 student tickets, and 25 child tickets.

Key Concepts

Linear Equation in Three Variables: A linear equation with three variables, where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form

$a x + b y + c z = d$

Every solution to the equation is an ordered triple, $(x, y, z)$ that makes the equation true.
How to solve a system of linear equations with three variables.
1. Write the equations in standard form
  If any coefficients are fractions, clear them.
2. Eliminate the same variable from two equations.
  Decide which variable you will eliminate.
  Work with a pair of equations to eliminate the chosen variable.
  Multiply one or both equations so that the coefficients of that variable are opposites.
  Add the equations resulting from Step 2 to eliminate one variable
3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
4. The two new equations form a system of two equations with two variables. Solve this system.
5. Use the values of the two variables found in Step 4 to find the third variable.
6. Write the solution as an ordered triple.
7. Check that the ordered triple is a solution to all three original equations.

Practice Makes Perfect

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In the following exercises, determine whether the ordered triple is a solution to the system.

${\begin{matrix} 2 x - 6 y + z = 3 \\ 3 x - 4 y - 3 z = 2 \\ 2 x + 3 y - 2 z = 3 \end{matrix}$

ⓐ $open parenthesis 3, 1, 3 close parenthesis$ ⓑ $open parenthesis 4, 3, 7 close parenthesis$

${\begin{matrix} - 3 x + y + z = −4 \\ - x + 2 y - 2 z = 1 \\ 2 x - y - z = −1 \end{matrix}$

ⓐ $open parenthesis −5, −7, 4 close parenthesis$ ⓑ $open parenthesis 5, 7, 4 close parenthesis$

ⓐ no ⓑ yes

${\begin{matrix} y - 10 z = −8 \\ 2 x - y = 2 \\ x - 5 z = 3 \end{matrix}$

ⓐ $open parenthesis 7, 12, 2 close parenthesis$ ⓑ $open parenthesis 2, 2, 1 close parenthesis$

${\begin{matrix} x + 3 y - z = 15 \\ y = \frac{2}{3} x - 2 \\ x - 3 y + z = −2 \end{matrix}$

ⓐ $(−6, 5, \frac{1}{2})$ ⓑ $(5, \frac{4}{3}, −3)$

ⓐ no ⓑ no

Solve a System of Linear Equations with Three Variables

In the following exercises, solve the system of equations.

${\begin{matrix} 5 x + 2 y + z = 5 \\ - 3 x - y + 2 z = 6 \\ 2 x + 3 y - 3 z = 5 \end{matrix}$

${\begin{matrix} 6 x - 5 y + 2 z = 3 \\ 2 x + y - 4 z = 5 \\ 3 x - 3 y + z = −1 \end{matrix}$

$open parenthesis 4, 5, 2 close parenthesis$

${\begin{matrix} 2 x - 5 y + 3 z = 8 \\ 3 x - y + 4 z = 7 \\ x + 3 y + 2 z = −3 \end{matrix}$

${\begin{matrix} 5 x - 3 y + 2 z = −5 \\ 2 x - y - z = 4 \\ 3 x - 2 y + 2 z = −7 \end{matrix}$

$open parenthesis 7, 12, −2 close parenthesis$

${\begin{matrix} 3 x - 5 y + 4 z = 5 \\ 5 x + 2 y + z = 0 \\ 2 x + 3 y - 2 z = 3 \end{matrix}$

${\begin{matrix} 4 x - 3 y + z = 7 \\ 2 x - 5 y - 4 z = 3 \\ 3 x - 2 y - 2 z = −7 \end{matrix}$

$open parenthesis −3, −5, 4 close parenthesis$

${\begin{matrix} 3 x + 8 y + 2 z = −5 \\ 2 x + 5 y - 3 z = 0 \\ x + 2 y - 2 z = −1 \end{matrix}$

${\begin{matrix} 11 x + 9 y + 2 z = −9 \\ 7 x + 5 y + 3 z = −7 \\ 4 x + 3 y + z = −3 \end{matrix}$

$open parenthesis 2, −3, −2 close parenthesis$

${\begin{matrix} \frac{1}{3} x - y - z = 1 \\ x + \frac{5}{2} y + z = −2 \\ 2 x + 2 y + \frac{1}{2} z = −4 \end{matrix}$

${\begin{matrix} x + \frac{1}{2} y + \frac{1}{2} z = 0 \\ \frac{1}{5} x - \frac{1}{5} y + z = 0 \\ \frac{1}{3} x - \frac{1}{3} y + 2 z = −1 \end{matrix}$

$open parenthesis 6, −9, −3 close parenthesis$

${\begin{matrix} x + \frac{1}{3} y - 2 z = −1 \\ \frac{1}{3} x + y + \frac{1}{2} z = 0 \\ \frac{1}{2} x + \frac{1}{3} y - \frac{1}{2} z = −1 \end{matrix}$

${\begin{matrix} \frac{1}{3} x - y + \frac{1}{2} z = 4 \\ \frac{2}{3} x + \frac{5}{2} y - 4 z = 0 \\ x - \frac{1}{2} y + \frac{3}{2} z = 2 \end{matrix}$

$open parenthesis 3, −4, −2 close parenthesis$

${\begin{matrix} x + 2 z = 0 \\ 4 y + 3 z = −2 \\ 2 x - 5 y = 3 \end{matrix}$

${\begin{matrix} 2 x + 5 y = 4 \\ 3 y - z = 3 \\ 4 x + 3 z = −3 \end{matrix}$

$open parenthesis −3, 2, 3 close parenthesis$

${\begin{matrix} 2 y + 3 z = −1 \\ 5 x + 3 y = −6 \\ 7 x + z = 1 \end{matrix}$

${\begin{matrix} 3 x - z = −3 \\ 5 y + 2 z = −6 \\ 4 x + 3 y = −8 \end{matrix}$

$open parenthesis −2, 0, −3 close parenthesis$

${\begin{matrix} 4 x - 3 y + 2 z = 0 \\ - 2 x + 3 y - 7 z = 1 \\ 2 x - 2 y + 3 z = 6 \end{matrix}$

${\begin{matrix} x - 2 y + 2 z = 1 \\ - 2 x + y - z = 2 \\ x - y + z = 5 \end{matrix}$

no solution

${\begin{matrix} 2 x + 3 y + z = 12 \\ x + y + z = 9 \\ 3 x + 4 y + 2 z = 20 \end{matrix}$

${\begin{matrix} x + 4 y + z = −8 \\ 4 x - y + 3 z = 9 \\ 2 x + 7 y + z = 0 \end{matrix}$

$x = \frac{203}{16}; y = \frac{–25}{16}; z = \frac{–231}{16};$

${\begin{matrix} x + 2 y + z = 4 \\ x + y - 2 z = 3 \\ - 2 x - 3 y + z = −7 \end{matrix}$

${\begin{matrix} x + y - 2 z = 3 \\ - 2 x - 3 y + z = −7 \\ x + 2 y + z = 4 \end{matrix}$

$(x, y, z)$ where $x = 5 z + 2; y = −3 z + 1; z$ is any real number

${\begin{matrix} x + y - 3 z = −1 \\ y - z = 0 \\ - x + 2 y = 1 \end{matrix}$

${\begin{matrix} x - 2 y + 3 z = 1 \\ x + y - 3 z = 7 \\ 3 x - 4 y + 5 z = 7 \end{matrix}$

$(x, y, z)$ where $x = 5 z - 2; y = 4 z - 3; z$ is any real number

Solve Applications using Systems of Linear Equations with Three Variables

In the following exercises, solve the given problem.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is twice the measure of the first angle. The third angle is twelve more than the second. Find the measures of the three angles.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is three times the measure of the first angle. The third angle is fifteen more than the second. Find the measures of the three angles.

45 degrees, 60 degrees, 75 degrees

After watching a major musical production at the theater, the patrons can purchase souvenirs. If a family purchases 4 t-shirts, the video, and 1 stuffed animal, their total is $135.

A couple buys 2 t-shirts, the video, and 3 stuffed animals for their nieces and spends $115. Another couple buys 2 t-shirts, the video, and 1 stuffed animal and their total is $85. What is the cost of each item?

The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of $65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of $140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 cans of popcorn for a total sales of $250. What is the cost of each item?

$20, $5, $10

Writing Exercises

In your own words explain the steps to solve a system of linear equations with three variables by elimination.

How can you tell when a system of three linear equations with three variables has no solution? Infinitely many solutions?

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first row contains the following statements: determine whether an ordered triple is a solution of a system of three linear equations with three variables, solve a system of linear equations with three variables, solve applications using systems of linear equations with three variables. The remaining columns are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?