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

So far we have worked with systems of two equations and two variables. Now we will extend that work to systems of three equations with three variables. But first let’s review what we already know about equations and systems involving up to two variables.

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. The points the lines have in common form the solution to the system. When we solve a system of two linear equations, there are three possible cases: the lines intersect in exactly one point (a consistent, independent system), the lines are parallel and never meet (an inconsistent system, no solution), or the two equations describe the same line, so every point on it is a solution (a consistent, dependent system, infinitely many solutions).

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.

All the points that are solutions to one such equation form a plane in three-dimensional space, and solving a system of three linear equations means finding what the three planes have in common. Just as with two variables, there are three possible cases:

• One solution. The three planes intersect at a single point. This is a consistent system with independent equations.
• No solution. The planes have no point in common — for example, they may be parallel, or two may be parallel while the third cuts across both, or each pair of planes may intersect in a different line so that no point lies on all three. This is an inconsistent system.
• Infinitely many solutions. The planes intersect along a common line (or are all the same plane), so every point on that line — or plane — is a solution. This is a consistent system with dependent equations.

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.

Example. Determine whether the ordered triple is a solution to the system:

(a) $(-2,-1,3)$ (b) $(-4,-3,4)$

(a) We substitute $x=-2$, $y=-1$, and $z=3$ into all three equations:

$(-2,-1,3)$ makes all three equations true, so it is a solution.

(b) We substitute $x=-4$, $y=-3$, and $z=4$ into all three equations:

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

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 gives 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 an original equation and find the third variable. We write our answer as an ordered triple and then check our results.

Example. Solve the system by elimination:

The system is easier to solve when each equation is already missing one of the variables.

Example. Solve the system by elimination:

Label the equations (1), (2), (3). Equation (1) is missing $y$, equation (2) is missing $x$, and equation (3) is missing $z$. We can eliminate $z$ from (1) and (2) by multiplying (2) by $2$ and adding it to (1):

Call this new equation (4). Now equations (3) and (4) both involve only $x$ and $y$:

Multiplying (3) by $-2$ and adding it to (4) eliminates $y$:

So $x=-4$. Substitute into (3) to find $y$:

Substitute $x=-4$ into (1) to find $z$:

The solution is $(-4,1,-3)$. Check that it makes all three original equations true:

When we solve a system and end up with no variables and a false statement, the system has no solution. It is inconsistent.

Example. Solve the system of equations:

Label the equations (1), (2), (3). Use (1) and (2) to eliminate $z$: multiplying (2) by $3$ and adding it to (1) gives

— call this (4). Use (2) and (3) to eliminate $z$ again: multiplying (2) by $2$ and adding it to (3) gives

— call this (5). Equations (4) and (5) have the same left side but different constants. Multiplying (5) by $-1$ and adding it to (4):

This is a false statement, so there is no solution. We are left with a false statement, which tells us the system is inconsistent.

When we solve a system and end up with no variables but a true statement, the system has infinitely many solutions. It is consistent with dependent equations, and the solution shows how two of the variables depend on the third.

Example. Solve the system of equations:

Label the equations (1), (2), (3). Use (1) and (3) to eliminate $x$: multiplying (1) by $-1$ and adding it to (3) gives

— call this (4). Use (1) and (2) to eliminate $x$ again: multiplying (1) by $-2$ and adding it to (2) gives

— call this (5). Use (4) and (5) to eliminate $y$: multiplying (4) by $-3$ and adding it to (5) gives

The true statement $0=0$ tells us the system is consistent with dependent equations — there are infinitely many solutions. Solve equation (4) for $y$:

Substitute $y=-2z+3$ into equation (1) to solve for $x$ in terms of $z$:

The solutions are all ordered triples of the form $(5z-5,\,-2z+3,\,z)$, where $z$ is any real number.

Solve Applications Using Systems of Linear Equations with Three Variables

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

Example. 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 organize the information in a chart:

“The number of student tickets sold is twice the number of adult tickets sold” translates to $y=2x$, or in standard form, $2x-y=0$. Write the system of equations:

Label the equations (1), (2), (3). Use (1) and (2) to eliminate $z$: multiplying (1) by $-8$ and adding it to (2) gives

— call this (4). Use (3) and (4) to eliminate $y$: multiplying (3) by $2$ and adding it to (4) gives

So $x=75$ — the department sold $75$ adult tickets. Substitute into (3) to find $y$:

So $150$ student tickets were sold. Substitute $x=75$ and $y=150$ into (1) to find $z$:

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

Key terms

linear equation in three variables — an equation of the form $ax+by+cz=d$, whose graph is a plane in three-dimensional space. ordered triple — a solution $(x,y,z)$ to an equation or system of equations in three variables. solutions of a system of linear equations with three variables — the ordered triples that make all three equations true; geometrically, the points common to all three planes. consistent system — a system with at least one solution. inconsistent system — a system with no solution; eliminating variables leaves a false statement like $0=-2$. dependent equations — equations whose graphs share a common line (or are the same plane), so the system has infinitely many solutions; eliminating variables leaves a true statement like $0=0$.

Practice

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

This section is adapted from Intermediate Algebra 2e, Section 4.4: Solve Systems of Equations with Three Variables by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: described the three-plane geometric cases (one solution, no solution, infinitely many solutions) in prose instead of recreating the 3D plane diagrams; recast the seven-step worked example as a markdown table; omitted the Be Prepared quiz, Self Check checklist, and media links; converted the practice problems (“Try Its”) into interactive exercises with instant feedback — including turning the no-solution and infinite-solutions Try Its into multiple-choice questions about the number of solutions, since a word answer can’t be graded by the math checker; and adapted selected end-of-section Practice Makes Perfect exercises into an interactive Practice block covering all three objectives.