Solve Systems of 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.

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.

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.

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.

How to Solve a System of Equations With Three Variables by Elimination

Solve the system by elimination: ${\begin{matrix} x - 2 y + z = 3 \\ 2 x + y + z = 4 \\ 3 x + 4 y + 3 z = −1 \end{matrix} .$

![The equations are x minus 2y plus z equals 3, 2x plus y plus z equals 4 and 3x plus 4y plus 3z equals minus 1. Step 1 is to write the equations in standard form. They are. If any coefficients are fractions, clear them. There are none.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005a_img.jpg) ![Step 2 is to eliminate the same variable from two equations. Decide which variable you will eliminate. We can eliminate the y’s from equations 1 and 2 by multiplying equation 2 by 2. 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. The new equation we get is 5x plus 3z equals 11.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005b_img.jpg) ![Step 3 is to repeat step 2 using two other equations and eliminate the same variable as in step 2. We can again eliminate the y’s using the equations 1, 3 by multiplying equation 1 by 2. Add the new equations and the result will be 5x plus 5z equals 5.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005c_img.jpg) ![Step 4. The two new equations form a system of two equations with two variables. Solve this system. Eliminating x, we get z equal to minus 3. Substituting this in one of the new equations, we get x equal to 4.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005d_img.jpg) ![Step 5 is to use the values of the two variables found in step 4 to find the third variable. Substituting values of x and z in one of the original equations, we get y equal to minus 1.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005e_img.jpg) ![Step 6 is to write the solution as an ordered triple 4, minus 1, minus 3.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005f_img.jpg) ![Step 7 is to check that the ordered triple is a solution to all three original equations. It makes all three equations true.](/algebra-intermediate-book/resources/CNX_IntAlg_Figure_04_04_005g_img.jpg)

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

{\begin{matrix} 3 x - 4 z = 0 (1) \\ 3 y + 2 z = −3 (2) \\ 2 x + 3 y = −5 (3) \end{matrix}

We can eliminate $z$

from equations (1) and (2) by multiplying equation (2) by 2 and then adding the resulting equations.

The equations are 3 x minus 4 equals 0, 3y plus 2 z equals minus 3 and 2 x plus 3 y equals minus 5. Multiply equation 2 by 2 and add to equation 1. We get 3 x plus 6 y equals minus 6. Notice that equations (3) and (4) both have the variables $x$

and $y$

. We will solve this new system for $x$

and $y$

Multiply equation 3 by minus 2 and add that to equation 4. We get x equal to minus 4. To solve for y, we substitute $x = −4$

into equation (3).

Substitute minus 4 into equation 3 and solve for y. We get y equal to 1. We now have $x = −4$

and $y = 1 .$

We need to solve for z. We can substitute $x = −4$

into equation (1) to find z.

Substituting minus 4 into equation 1 for x, we get z equal to minus 3. We write the solution as an ordered triple. $(−4, 1, −3)$

We check that the solution makes all three equations true.

\begin{matrix} \begin{matrix} 3 x - 4 z & = & 0 (1) \\ 3 (−4) - 4 (−3) & \overset{?}{=} & 0 \\ 0 & = & 0 ✓ \end{matrix} & \begin{matrix} 3 y + 2 z & = & −3 (2) \\ 3 (1) + 2 (−3) & \overset{?}{=} & - 3 \\ −3 & = & −3 ✓ \end{matrix} & \begin{matrix} \begin{matrix} 2 x + 3 y & = & −5 (3) \\ 2 (−4) + 3 (1) & \overset{?}{=} & - 5 \\ −5 & = & −5 ✓ \end{matrix} \\ The solution is (−4, 1, −3) . \end{matrix} \end{matrix}

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.

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.

The equations are x plus 2y minus z equals 1, 2x plus 7y plus 4z equals 11 and x plus 3y plus z equals 4. Multiply equation 1 with minus 1 and add it to equation 3. We get equation 4, y plus 2z equals 3. Use equation (1) and (2) to eliminate x again.

Multiply equation 1 with minus 2 and add it to equation 2. We get equation 5, 3y plus 6z equals 9. 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 $0 = 0$

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.

Key Concepts

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}

ⓐ $(3, 1, 3)$

ⓑ $(4, 3, 7)$

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

ⓐ $(−5, −7, 4)$

ⓑ $(5, 7, 4)$

ⓐ no ⓑ yes

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

ⓐ $(7, 12, 2)$

ⓑ $(2, 2, 1)$

{\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 ⓑ yes

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}

(4, 5, 2)

{\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}

(7, 12, −2)

{\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}

(−3, −5, 4)

{\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}

(2, −3, −2)

{\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}

(6, −9, −3)

{\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}

(3, −4, −2)

{\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}

(−3, 2, 3)

{\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}

(−2, 0, −3)

{\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}

no solution

{\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 if 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 the measure if the first angle. The third angle is fifteen more than the second. Find the measures of the three angles.

42, 50, 58

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 animal 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 can 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.

ⓑ 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?

We will use a chart to organize the information.
Number of students is twice number of adults.
Rewrite the equation in standard form.	$\begin{matrix} y & = & 2 x \\ 2 x - y & = & 0 \end{matrix}$

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

Use (3) and (4) to eliminate $y .$

Solve for x.	$x = 75$ adult tickets
Use equation (3) to find y.	$−2 x + y = 0$
Substitute $x = 75 .$	$\begin{matrix} −2 (75) + y & = & 0 \\ −150 + y & = & 0 \\ y & = & 150 student tickets \end{matrix}$
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.