Solve Systems of Equations with Three Variables

4.4 Solve Systems of Equations with Three Variables

Topics covered in this section are:

4.4.1 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, $ax+by=c$, 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.

The 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 $ax+by+cz=d$, every solution to the equation is an ordered triple $(x, y, z)$, that makes the equation true.

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

$ax+by+cz=d$

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.

This figure shows three intersecting planes with one point in common. It is labeled Consistent system and Independent equations.
This figure has three parallel planes with no points in common. It is labeled Inconsistent system.
This figure shows two planes that are coincident and parallel to the third plane. The planes have no points in common.
In this figure, the two planes are parallel and each intersects the third plane. The planes have no points in common.
In this figure, each plane intersects the other two, but all three share no points. The planes have no points in common.
In this figure, three planes intersect in one line. There is just one line, so there are infinitely many solutions.
In this 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 this 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.

SOLUTIONS OF A SYSTEM OF LINEAR EQUATIONS WITH THREE VARIABLES

Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple $(x, y, z)$.

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.

Example 1

Determine whether the ordered triple is a solution to the system:
$\Bigg\{ \begin{align*} &x-y+z=2 \\ &2x-y-z=-6 \\ &2x+2y+z=-3 \end{align*}$

  • $(-2, -1, 3)$
  • $(-4, -3, 4)$
Solution

Part 1

$\Bigg\{ \begin{align*} &x-y+z=2 \\ &2x-y-z=-6 \\ &2x+2y+z=-3 \end{align*}$

We substitute $x=\textcolor{red}{-2}$ and $y=\textcolor{blue}{-1}$ and $z=\textcolor{green}{3}$ into all three equations.

$\tiny \begin{align*} x-y+z&=2\\ \textcolor{red}{-2}-(\textcolor{blue}{-1})+\textcolor{green}{3}&=2 \\ 2 &=2 \ \checkmark \end{align*}$$ \tiny \begin{align*} 2x-y-z&=-6\\ 2 \cdot (\textcolor{red}{-2})-(\textcolor{blue}{-1})-\textcolor{green}{3}&=-6 \\ -6 &=-6 \ \checkmark \end{align*}$$\tiny \begin{align*} 2x+2y+z&=-3\\ 2 \cdot (\textcolor{red}{-2})+2(\textcolor{blue}{-1})+\textcolor{green}{3}&=-3 \\ -3 &=-3 \ \checkmark \end{align*}$

$(-2, -1, 3)$ does make all three equations true. So, $(-2, -1, 3)$ is a solution.

Part 2

$\Bigg\{ \begin{align*} &x-y+z=2 \\ &2x-y-z=-6 \\ &2x+2y+z=-3 \end{align*}$

We substitute $x=\textcolor{red}{-4}$ and $y=\textcolor{blue}{-3}$ and $z=\textcolor{green}{4}$ into all three equations.

$\tiny \begin{align*} x-y+z&=2\\ \textcolor{red}{-4}-(\textcolor{blue}{-3})+\textcolor{green}{4}&=2 \\ 3 &≠2 \end{align*}$$\tiny \begin{align*} 2x-y-z&=-6\\ 2 \cdot (\textcolor{red}{-4})-(\textcolor{blue}{-3})-\textcolor{green}{4}&=-6 \\ -9 &≠-6 \end{align*}$$\tiny \begin{align*} 2x+2y+z&=-3\\ 2 \cdot (\textcolor{red}{-4})+2(\textcolor{blue}{-3})+\textcolor{green}{4}&=-3 \\ -10 &≠-3 \ \checkmark \end{align*}$

$(-4, -3, 4)$ does not make all three equations true. So, $(-4, -3, 4)$ is not a solution.

4.4.2 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.

Example 2

Solve the system by elimination: $\Bigg\{ \begin{align*} &x-2y+z=3\\ &2x+y+z=4 \\ &3x+4y+3z=-1 \end{align*}$

Solution
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.
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.
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.
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.
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.
Step 6 is to write the solution as an ordered triple 4, minus 1, minus 3.
Step 7 is to check that the ordered triple is a solution to all three original equations. It makes all three equations true.

The steps are summarized here.

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.

Example 3

Solve: $\Bigg\{ \begin{align*} 3x-4x&=0\\ 3y+3z&=0 \\ 2x+3y&=-5 \end{align*}$

Solution

$\Bigg\{ \begin{align*} 3x-4x&=0\ \ \ \ \ (1) \\ 3y+3z&=0\ \ \ \ \ (2) \\ 2x+3y&=-5\ \ (3) \end{align*}$

We can eliminate $z$ from equations (1) and (2) by multiplying equation (2) by $2$ 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).

$\begin{align*} 2x+3y&=-5 \\ 2(\textcolor{red}{-4})+3y&=-5 \\ -8+3y&=-5 \\3y&=3\\ y&=1 \end{align*}$

We now have $x=-4$ and $y=1$. We need to solve for $z$. We can substitute $x=-4$ into equation (1) to find $z$.

$\begin{align*} 3x-4z&=0 \\ 3(\textcolor{red}{-4})-4z&=0 \\ -12-4z&=0 \\-4z&=12\\ z&=-3 \end{align*}$

We write the solution as an ordered triple. $(-4, 1, -3)$

We check that the solution makes all three equations true.

$\tiny \begin{align*} 3x-4z&=0\\ 3(-4)-4(-3)&=0 \\ 0 &=0 \ \checkmark \end{align*}$$ \tiny \begin{align*} 3y+2z&=-3\\ 3(1)+2(-3)&=-3 \\ -3 &=-3 \ \checkmark \end{align*}$$\tiny \begin{align*} 2x+3y&=-5\\ 2(-4)+3(1)&=-5 \\ -5 &=-5 \ \checkmark \end{align*}$

The solution is $(-4, 1, -3)$.

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.

Example 4

Solve the system of equations: $\Bigg\{ \begin{align*} &x+2y-3z=-1 \\ &x-3y+z=1 \\ &2x-y-2z=2 \end{align*}$

Solution

$\Bigg\{ \begin{align*} &x+2y-3z=-1 \ \ (1) \\ &x-3y+z=1 \ \ \ \ \ \ \ (2) \\ &2x-y-2z=2 \ \ \ \ \ (3) \end{align*}$

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

The equations are x plus 2y minus 3z equals minus 1, x minus 3y plus z equals 1 and 2x minus y minus 2z equals 2.

Use equations (2) and (3) to eliminate $z$ again.

Multiplying equation 2 by 3 and adding it to equation 1, we get equation 4, 4x minus 7y equals 2. Multiplying equation 2 by 2 and adding it to equation 3, we get equation 5, 4x minus 7y equals 4.

Use equations (4) and (5) to eliminate a variable.

Equations 4 and 5 both have 2 variables. Multiply equation 5 by minus 1 and add it to equation 4. We get 0 equal to minus 2, which is false.

There is no solution.

We are left with a false statement and this tells us the system is inconsistent and has no solution.

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: $\Bigg\{ \begin{align*} &x+2y-z=1\\&2x+7y+4z=11 \\ &x+3y+z=4 \end{align*}$

Solution

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

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

Multiply equation 4 with minus 3 and add it to equation 5. We get 0 equal to 0. There are infinite many solutions. Solving equation 4 for y, we get y equal to minus 2z plus 3. Substituting this into equation 1, we get x equal to 5z minus 5. The true statement 0 equal to 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 is 5z minus 5, y is minus 2z plus 3 and z is any real number."
There are infinitely many solutions.
Solve equations (4) for $y$.Represent the solution showing how $x$ and $y$ are dependent on $z$.
$\begin{align*} y+2z&=3\\ y&=-2x+3 \end{align*}$
Use equation (1) to solve for $x$. $x+2y-z-1$
Substitute $y=-2x+3$. $\begin{align*} x+2(-2z+3)-z&=1\\ x-4z+6-z&=1\\x-5z+6&=1\\ x&=5z-5 \end{align*}$

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 $x=5z−5$; $y=−2z+3$ and $z$ is any real number.

4.4.3 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?

Solution
We will use a chart to organize the information.
The number of students is twice the number of adults.
Rewrite the equation in standard form.$y=2x$
$2x-y=0$
Write the system of equations.$\Bigg\{ \begin{align*} x+y+z&=250\ \ \ \ \ \ (1) \\15x+10y+8z&=2825\ \ \ \ (2) \\ -2x+y \ \ \ \ \ \ \ &=0 \ \ \ \ \ \ \ \ \ \ (3)\end{align*}$
Use equations (1) and (2) to eliminate $z$.
Use equations (3) and (4) to eliminate $y$.
Solve for $x$.$x=75$ adult tickets
Use equation (3) to find $y$. Substitute $x=75$.$\begin{align*} -2x+y&=0 \\-2(75)+y&=0 \\ -150+y&=0\\ y&=150 \text{ student tickets} \end{align*}$
Use equation (1) to find $z$. Substitute in the values $x=75$ and $y=150$.$\begin{align*} x+y+z&=250\\ 75+150+z&=250 \\225+z&=250\\ z&=25 \text{ child tickets} \end{align*}$
Write the solution.The theater department sold $75$ adult tickets, $150$ student tickets, and $25$ child tickets.
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