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Study Guides > Intermediate Algebra

Read: Systems of Three Equations in Three Variables

Learning Objectives

  • Define the solution to a system of three equations in three variables
  • Determine whether an ordered triple is a solution to a system
  • Solve a systems of equations in three variables using elimination and back-substitution
The solution set to a system of three equations in three variables is an ordered triple [latex]\left(x,y,z\right)[/latex]. Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes.

Solution Set, One Solution

The figure below illustrates how a system with three variables can have one solution. Systems that have a single solution are those which result in a solution set consisting of an ordered triple [latex]\left(x,y,z\right)[/latex]. Graphically, the ordered triple defines a point that is the intersection of three planes in space. Screen Shot 2016-07-11 at 2.08.59 PM
In the first example, we will determine whether an ordered triple is a solution for a systems of three linear equations in three variables.

Example

Determine whether the ordered triple [latex]\left(3,-2,1\right)[/latex] is a solution to the system.
[latex]\begin{array}{l}\text{ }x+y+z=2\hfill \\ 6x - 4y+5z=31\hfill \\ 5x+2y+2z=13\hfill \end{array}[/latex]

Answer: We will check each equation by substituting in the values of the ordered triple for [latex]x,y[/latex], and [latex]z[/latex]. [latex-display]\begin{array}{ccccc}\begin{array}{r}\hfill x+y+z=2\\ \hfill \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}6x - 4y+5z=31\\ \hfill 6\left(3\right)-4\left(-2\right)+5\left(1\right)=31\\ \hfill 18+8+5=31\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}5x+2y+2z=13\\ \hfill 5\left(3\right)+2\left(-2\right)+2\left(1\right)=13\\ \hfill \text{}15 - 4+2=13\\ \hfill \text{True}\end{array}\end{array}[/latex-display] The ordered triple [latex]\left(3,-2,1\right)[/latex] is indeed a solution to the system.

How To: Given a linear system of three equations, solve for three unknowns.

  1. Pick any pair of equations and solve for one variable.
  2. Pick another pair of equations and solve for the same variable.
  3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
  4. Back-substitute known variables into any one of the original equations and solve for the missing variable.
Solving a system with three variables is very similar to solving one with two variables.  It is important to keep track of your work as the addition of one more equation can create more steps in the solution process. In the example that follows we will solve the system by first using the elimination method to solve for x, then using back-substitution.

Example

Solve the system [latex-display]\displaystyle\begin{cases}x-\dfrac{1}{3}y+\dfrac{1}{2}z=1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,y-\dfrac{1}{2}z=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z=-1\end{cases}[/latex-display]

Answer: The third equation states that [latex]z = −1[/latex], so we substitute this into the second equation to obtain a solution for [latex]y[/latex]. [latex-display]\begin{array}y-\dfrac{1}{2}(-1)=4\\y+\dfrac{1}{2}=4\\y=4-\dfrac{1}{2}\\y=\dfrac{8}{2}-\dfrac{1}{2}\\y=\dfrac{7}{2}\end{array}[/latex-display] Now we have two of our solutions and we can substitute them both into the first equation to solve for [latex]x[/latex]. [latex-display]\begin{array}x-\dfrac{1}{3}\left(\dfrac{7}{2}\right)+\dfrac{1}{2}\left(-1\right)=1\\x-\dfrac{7}{6}-\dfrac{1}{2}=1\\x-\dfrac{7}{6}-\dfrac{3}{6}=1\\x-\dfrac{10}{6}=1\\x=1+\dfrac{10}{6}\\x=\dfrac{6}{6}+\dfrac{10}{6}\\x=\dfrac{16}{6}=\dfrac{8}{3}\end{array}[/latex-display] Now we have our ordered triple, remember that where you place the solutions matters!

Answer

[latex-display](x,y,z)=\left(\dfrac{8}{3},\dfrac{7}{2},-1\right)[/latex-display]

Analysis of the solution:

Each of the lines in this system represents a plane (think about a sheet of paper). If you imagine three sheets of notebook paper each representing a portion of these planes, you will start to see the complexities involved in how three such planes can intersect. Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line, so our intersection point is where all three of these lines meet.
Three Planes Intersecting. Three Planes Intersecting.
In the following video we show another example of using back-substitution to solve a system in three variables. https://youtu.be/HHIjTChrIxE In the next example, we will not start with a solution, but will need to use the method of elimination to find our first solution.

Example

Find a solution to the following system:
[latex]\begin{array}{lll}x-y+z=5\,\,\,\,(1)\\-2y+z=6\,\,\,\,(2)\\2y-2z=-12\,\,\,\,(3)\end{array}[/latex]

Answer:

We labeled the equations this time to be able to keep track of things a little more easily. The most obvious first step here is to eliminate [latex]y[/latex] by adding equations (2) and (3).
[latex]\begin{array}{lll}\,\,\,\,\,\,\,\,\,-2y+z=6\,\,\,\,(2)\\\underline{+\,\,\,\,2y-2z=-12}\,\,\,\,(3)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-z=-6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z=6\end{array}[/latex]
  Now we can substitute the value for [latex]z[/latex] that we obtained into equation [latex](2)[/latex].

[latex]\begin{array}{rrr}-2y+(6)=6\\-2y=6-6\\-2y=0\\\,\,\,\,y=0\end{array}[/latex]

Be careful here not to get confused with a solution of [latex]y = 0[/latex] and an inconsistent solution.  It's ok for variables to equal [latex]0[/latex]. Now we can substitute [latex]z = 6[/latex] and [latex]y = 0[/latex] back into the original equation.

[latex]\begin{array}{rrr}x-y+z=5\\x-0+6=5\\x+6=5\\x=5-6\\x=-1\end{array}[/latex]

Answer

[latex-display](x,y,z)=(-1,0,6)[/latex-display]

In the following videos we show more examples of the algebra you may encounter when solving systems with three variables. https://youtu.be/r6htz3gaHZ0 https://youtu.be/3RbVSvvRyeI

Summary

  • The solution to a system of linear equations in three variables is an ordered triple in the form [latex](x,y,z)[/latex]
  • Solutions can be verified using substitution and the order of operations
  • Systems of three variables can be solved using the same techniques as we used to solve systems with two variables, including elimination and substitution.

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • College Algebra: 8.1 Systems of Linear Equations: Gaussian Elimination. Authored by: Stitz, Carl and Zeager, Jeff. Located at: https://www.stitz-zeager.com/szca07042013.pdf. License: CC BY: Attribution.
  • Ex: Solve a System of 3 Equations with 3 Unknowns Using Back Substitution. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex 2: System of Three Equations with Three Unknowns Using Elimination. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex 1: System of Three Equations with Three Unknowns Using Elimination. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.

CC licensed content, Specific attribution