7.2: Systems of Linear Equations - Three Variables

Learning Objectives Solve systems of three equations in three variables. Identify inconsistent systems of equations containing three variables. Express the solution of a system of dependent equations containing three variables.

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Figure \( \PageIndex{ 1 } \)

(credit: "Elembis," Wikimedia Commons)
 

John received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested $4,000 more in municipal funds than in municipal bonds. He earned $670 in interest the first year. How much did John invest in each type of fund?

Understanding the correct approach to setting up problems such as the one above makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics.

Solving Systems of Three Equations in Three Variables

In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian Elimination, named after the prolific German mathematician Carl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution \(( x,y,z )\), which we call an ordered triple.

Definition: Upper Triangular Form

A \( 3 \times 3 \) system is said to be in upper triangular form if it is written as follows:\[ \begin{cases} Ax & + & By & + & Cz & = & D \\ & & Ey & + & Fz & = & G \\ & & & & Hz & = & K \\ \end{cases} \nonumber \]The upper triangular form for an \( n \times n \) systems is likewise defined to be a system in the form\[ \begin{cases}
a_{11} x_1 & + & a_{12} x_2 & + & a_{13} x_3 & + & \cdots & + & a_{1n} x_n & = & c_1 \\
                 &    & a_{22} x_2 & + & a_{23} x_3 & + & \cdots & + & a_{2n} x_n & = & c_2 \\
                 &    &                  &    & a_{33} x_3 & + & \cdots & + & a_{3n} x_n & = & c_3 \\
                 &    &                  &    &                  & \ddots &    &    &          \vdots         &   & \vdots \\
                 &    &                  &     &                 &           & \ddots &     &    \vdots         &   & \vdots \\
                &     &                  &     &                  &           &          & &   a_{nn} x_n &  = & c_n \\
 \end{cases} \nonumber \]

There are entire branches of Mathematics dedicated to developing faster methods for solving large systems; however, in this section, we limit ourselves to \( 3 \times 3 \) systems. Therefore, let's focus on the upper triangular form for a \( 3\times 3 \) system.

The upper triangular form is useful because, in the \( 3 \times 3 \) case, the third equation can be solved for \(z\), and then we back-substitute to find \(y\) and \(x\). Therefore, our task is to find a method to write a given system in upper triangular form. To achieve our goal, we introduce the following theorem.

Theorem: Invariant Solutions Under Row Operations

The solution to a system of linear equations in invariant (remains unchanged) under the following operations:

1. Row Exchange: Interchange the order of any two equations.
2. Row Scaling: Multiply both sides of an equation by a nonzero constant.
3. Row Replacement: Add a nonzero multiple of one equation to another equation.

As mentioned in the previous section, the solution set to a three-by-three system is an ordered triple \(\{ ( x,y,z ) \} \). 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.

Theorem: Number of Solutions to 3x3 Linear Systems

A system of three linear equations involving three unknowns has either one, none, or infinitely many solutions.

Figures \( \PageIndex{ 2 } \) and \( \PageIndex{ 3 } \) illustrate possible solution scenarios for three-by-three systems.

Figure \( \PageIndex{ 2 } \)

(a): Three planes intersect at a single point, representing a three-by-three system with a single solution.
(b): Three planes intersect in a line, representing a three-by-three system with infinite solutions.
 

Figure \( \PageIndex{ 3 } \)

All three figures represent three-by-three systems with no solution.
(a): The three planes intersect with each other, but not at a common point.
(b): Two of the planes are parallel and intersect with the third plane, but not with each other.
(c): All three planes are parallel, so there is no point of intersection.
 

• Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple \(\{ ( x,y,z ) \} \). Graphically, the ordered triple defines a point that is the intersection of three planes in space.
• Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as \(0=0\). Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space.
• Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as \(3=0\). Graphically, a system with no solution is represented by three planes with no point in common.

Example \( \PageIndex{ 1 } \): Determining Whether an Ordered Triple Is a Solution to a System

Determine whether the ordered triple \(( 3,−2,1 )\) is a solution to the system.\[ \begin{cases} x & + & y & + & z & = & 2 \\ 6x & − & 4y & + & 5z & = & 31 \\ 5x & + & 2y & + & 2z & = & 13 \\ \end{cases} \nonumber \]

Solution

We will check each equation by substituting in the values of the ordered triple for \(x\), \(y\), and \(z\).\[ \begin{array}{c|cc|cc} \begin{array}{rrcl} & x + y + z & = & 2 \\[6pt] \implies & (3)+(−2)+(1) & = & 2 \\[6pt] \implies & 2 & = & 2 \checkmark \\[6pt] \end{array} & \quad & \begin{array}{rrcl} & 6x−4y+5z & = & 31 \\[6pt] \implies & 6(3)−4(−2)+5(1) & = & 31 \\[6pt] \implies & 18+8+5 & = & 31 \\[6pt] \implies & 31 & = & 31 \checkmark \\[6pt] \end{array} & \quad & \begin{array}{rrcl} & 5x+2y+2z & = & 13 \\[6pt] \implies & 5(3)+2(−2)+2(1) & = & 13 \\[6pt] \implies & 15−4+2 & = & 13 \\[6pt] \implies & 13 & = & 13 \checkmark \\[6pt] \end{array} \end{array} \nonumber \]The ordered triple \(( 3,−2,1 )\) is indeed a solution to the system.

The process of solving a linear system of three equations involving three unknowns can be cumbersome at times - especially if you do not have a systematic approach. We will begin by taking the following approach, which employs the Substitution Method:

1. Pick a variable and solve one of the equations for that variable.
2. Substitute this solution into the remaining two equations.
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.

It is important to note that the method we have written is not ideal, but it links well with previously-learned methods. We will introduce a more algorithmic approach in a moment.

Example \( \PageIndex{ 2 } \): Solving a System of Three Equations in Three Variables by Substitution

Find a solution to the following system:\[ \begin{cases} x & − & 2y & + & 3z & = & 9 \\ −x & + & 3y & − & z & = & −6 \\ 2x & − & 5y & + & 5z & = & 17 \\ \end{cases} \nonumber \]

Solution

We begin by choosing a variable to solve for. While we could choose any of the three variables, \( x \) is easy to solve for in the first equation, so that will be our choice.\[ x - 2y + 3z = 9 \implies x = 9 + 2y - 3z. \nonumber \]Substituting this for \( x \) in the remaining two equations, we get the following:\[ \begin{array}{ccc} \begin{array}{rrcl} & -x + 3y - z & 2 = -6 \\[6pt] \implies & -(9 + 2y - 3z) + 3y - z & = & -6 \\[6pt] \implies & -9 - 2y + 3z + 3y - z & = & -6 \\[6pt] \implies & -9 + y + 2z & = & -6 \\[6pt] \implies & y + 2z & = & 3 \\[6pt] \end{array} & \quad \text{and} \quad & \begin{array}{rrcl} & 2x - 5y + 5z & = & 17 \\[6pt] \implies & 2(9 + 2y - 3z) - 5y + 5z & = & 17 \\[6pt] \implies & 18 + 4y - 6z - 5y + 5z & = & 17 \\[6pt] \implies & 18 - y - z & = & 17 \\[6pt] \implies & - y - z & = & -1 \\[6pt] \end{array} \end{array} \nonumber \]At this point, we have reduced the system of three equations in three unknowns down to the following system of two linear equations in two unknowns.\[ \begin{cases} y & + & 2z & = & 3 \\ -y & - & z & = & -1 \\ \end{cases} \nonumber \]We can solve this \(2 \times 2\) system using the Elimination Method. Adding the two equations, we get\[ \begin{array}{rrclcl} & y & + & 2z & = & 3 \\[6pt] + & -y & - & z & = & -1 \\[6pt] \hline & & & z & = & 2 \\[6pt] \end{array} \nonumber \]Substituting \( z = 2 \) into \( y + 2z = 3 \), we get\[ y + 2(2) = 3 \implies y = -1. \nonumber \]Finally, going back to our original equation solved for \( x \), we let \( z = 2 \) and \( y = -1 \):\[ \begin{array}{rrclcl} & x & = & 9 + 2y - 3z & & \\[6pt] \implies & x & = & 9 + 2(-1) - 3(2) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & x & = & 9 -2 - 6 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & x & = & 1 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]The solution is the ordered triple \(( 1,−1,2 )\). See Figure \( \PageIndex{ 4 } \).

Figure \( \PageIndex{ 4 } \)
 

Solving linear systems using the method applied in Example \( \PageIndex{ 2 } \) can get us into trouble when dealing with larger systems. Therefore, it is beneficial to showcase a method based on the Elimination Method we learned previously. The process is as follows:

1. Pick any pair of equations and use the technique discussed in the previous section to eliminate one variable.
2. Pick another pair of equations and eliminate 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.

Example \( \PageIndex{ 3 } \): Solving a System of Three Equations in Three Variables by Elimination

Find a solution to the following system:\[ \begin{cases} x & − & 2y & + & 3z & = & 9 & \quad & (E1) \\ −x & + & 3y & − & z & = & −6 & \quad & (E2) \\ 2x & − & 5y & + & 5z & = & 17 & \quad & (E3) \\ \end{cases} \nonumber \]

Solution

When learning the Elimination Method, it's best to label the equations (in this case, we used \((E1)\), \((E2)\), and \((E3)\)). We begin by choosing a variable to eliminate. The ideal looks to be \( x \). We will add \((E1)\) to \((E2)\) to eliminate the \(x\) from those equations.\[ \begin{array}{crccccccc} & (E1): & x & - & 2y & + & 3z & = & 9 \\[6pt] + & (E2): & -x & + & 3y & - & z & = & -6 \\[6pt] \hline & & & & y & + & 2z & = & 3 \\[6pt] \end{array} \nonumber \]We now and \(2 \times (E2)\) to \((E3)\) to eliminate the \(x\) from those.\[ \begin{array}{crccccccc} & 2 \times (E2): & -2x & + & 6y & - & 2z & = & -12 \\[6pt] + & (E3): & 2x & - & 5y & + & 5z & = & 17 \\[6pt] \hline & & & & y & + & 3z & = & 5 \\[6pt] \end{array} \nonumber \]At this point, we have reduced the system of three equations in three unknowns down to the following system of two linear equations in two unknowns.\[ \begin{cases} y & + & 2z & = & 3 & \quad & (E4) \\ y & + & 3z & = & 5 & \quad & (E5) \\ \end{cases} \nonumber \]We choose to continue with the Elimination Method by adding the opposite of \( (E4) \) to \( (E5) \).\[ \begin{array}{crccccc} & -(E4): & -y & - & 2z & = & -3 \\[6pt] + & (E5): & y & + & 3z & = & 5 \\[6pt] \hline & & & & z & = & 2 \\[6pt] \end{array} \nonumber \]Substituting \( z = 2 \) into \( y + 2z = 3 \), we get\[ y + 2(2) = 3 \implies y = -1. \nonumber \]Finally, we can use any of the equations involving \(x\) and we let \( z = 2 \) and \( y = -1 \). In this case, let's just use \( (E1) \):\[ \begin{array}{rrclcl} & x - 2y + 3z & = & 9 & & \\[6pt] \implies & x - 2(-1) + 3(2) & = & 9 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & x + 2 + 6 & = & 9 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & x + 8 & = & 9 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & x & = & 1 & \quad & \left( \text{subtracting }8\text{ from both sides} \right) \\[6pt] \end{array} \nonumber \]The solution is the ordered triple \(( 1,−1,2 )\).

Checkpoint \( \PageIndex{ 3 } \)

Solve the system of equations in three variables.\[ \begin{cases} 2x & + & y & − & 2z & = & −1 \\ 3x & − & 3y & − & z & = & 5 \\ x & − & 2y & + & 3z & = & 6 \\ \end{cases} \nonumber \]

Example \( \PageIndex{ 4 } \): Solving a Real-World Problem Using a System of Three Equations in Three Variables

In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund?

Solution

To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:\[ \begin{array}{rcl} x & = & \text{amount invested in money-market fund} \\ y & = & \text{amount invested in municipal bonds} \\ z & = & \text{amount invested in mutual funds} \\ \end{array} \nonumber \]The first equation indicates that the sum of the three principal amounts is $12,000.\[x+y+z=12,000 \nonumber \]We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.\[z=y+4,000 \nonumber \]The third equation shows that the total amount of interest earned from each fund equals $670.\[0.03x+0.04y+0.07z=670 \nonumber \]Then, we write the three equations as a system.\[ \begin{cases} x & + & y & + & z & = & 12,000 \\ & − & y & + & z & = & 4,000 \\ 0.03x & + & 0.04y & + & 0.07z & = & 670 \\ \end{cases} \nonumber \]To make the calculations simpler, we can multiply the third equation by 100. Thus,\[ \begin{cases} x & + & y & + & z & = & 12,000 & \quad & (E1) \\ & − & y & + & z & = & 4,000 & \quad & (E2) \\ 3x & + & 4y & + & 7z & = & 67,000 & \quad & (E3) \\ \end{cases} \nonumber \]Since one of the equations is already missing a term involving \(x\), let's eliminate the \(x\) in the other two equations. This requires us to add \(-3 \times (E1)\) to \( (E3)\).\[ \begin{array}{crccccccc} & -3 (E1): & -3x & - & 3y & - & 3z & = & -36,000 \\[6pt] + & (E3): & 3x & + & 4y & + & 7z & = & 67,000 \\[6pt] \hline & & & & y & + & 4z & = & 31,000 \\[6pt] \end{array} \nonumber \]We can take this result and place it in a new system involving \((E2)\) - this is because \((E2)\) is already missing a term with \(x\).\[ \begin{cases} -y & + & z & = & 4,000 & \quad & (E2) \\ y & + & 4z & = & 31,000 & \quad & (E4) \\ \end{cases} \nonumber \]Adding these two equations together immediately eliminates \( y \) to get\[ 5z = 35,000 \implies z = 7,000. \nonumber \]We now substitute this into \((E2)\) to get\[ -y + 7,000 = 4,000 \implies -y = -3,000 \implies y = 3,000. \nonumber \]Finally, we substitute these values into any equation involving \( x \), \( y \), and \( z \). Let's choose \( (E1) \):\[ x + 3,000 + 7,000 = 12,000 \implies x = 2,000. \nonumber \]John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.

Identifying Inconsistent Systems of Equations Containing Three Variables

Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The Elimination Method will result in a false statement, such as \(3=7\) or some other contradiction.

Example \( \PageIndex{ 5 } \): Solving an Inconsistent System of Three Equations in Three Variables

Solve the following system.\[ \begin{cases} x & − & 3y & + & z & = & 4 & \quad & (E1) \\ −x & + & 2y & − & 5z & = & 3 & \quad & (E2) \\ 5x & − & 13y & + & 13z & = & 8 & \quad & (E3) \\ \end{cases} \nonumber \]

Solution

Looking at the coefficients of \(x\), we can see that we can eliminate \(x\) by adding equation \( (E1) \) to equation \( (E2) \).\[ \begin{array}{crccccccc} & (E1): & x & − & 3y & + & z & = & 4 \\[6pt] + & (E2): & −x & + & 2y & − & 5z & = & 3 \\[6pt] \hline & & & & -y & − & 4z & = & 7 \\[6pt] \end{array} \nonumber \]Next, we multiply equation \( (E1) \) by \(−5\) and add it to equation \( (E3) \).\[ \begin{array}{crccccccc} & -5(E1): & -5x & + & 15y & - & 5z & = & -20 \\[6pt] + & (E3): & 5x & - & 13y & + & 13z & = & 8 \\[6pt] \hline & & & & 2y & + & 8z & = & -12 \\[6pt] \end{array} \nonumber \]We now have the simpler system\[ \begin{cases} -y & − & 4z & = & 7 & \quad & (E4) \\ 2y & + & 8z & = & -12 & \quad & (E5) \\ \end{cases} \nonumber \]We now multiply equation \( (E4) \) by 2 and add it to equation \( (E5) \).\[ \begin{array}{crccccc} & 2(E4): & -2y & - & 8z & = & 14 \\[6pt] + & (E5): & 2y & + & 8z & = & -12 \\[6pt] \hline & & & & 0 & = & 2 \\[6pt] \end{array} \nonumber \]The final equation, \(0=2\), is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent.

Checkpoint \( \PageIndex{ 5 } \)

Solve the system of three equations in three variables.\[ \begin{cases} x & + & y & + & z & = & 2 \\ & & y & − & 3z & = & 1 \\ 2x & + & y & + & 5z & = & 0 \\ \end{cases} \nonumber \]

Expressing the Solution of a System of Dependent Equations Containing Three Variables

We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line.

Example \( \PageIndex{ 6 } \): Finding the Solution to a Dependent System of Equations

Find the solution to the given system of three equations in three variables.\[ \begin{cases} 2x & + & y & − & 3z & = & 0 & \quad & (E1) \\ 4x & + & 2y & − & 6z & = & 0 & \quad & (E2) \\ x & − & y & + & z & = & 0 & \quad & (E3) \\ \end{cases} \nonumber \]

Solution

First, we can multiply equation \( (E1) \) by \(−2\) and add it to equation \( (E2) \).\[ \begin{array}{crccccccc} & -2(E1): & -4x & − & 2y & + & 6z & = & 0 \\[6pt] + & (E2): & 4x & + & 2y & − & 6z & = & 0 \\[6pt] \hline & & & & & & 0 & = & 0 \\[6pt] \end{array} \nonumber \]We do not need to proceed any further. The result we get is an identity, \(0=0\), which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation \( (E3) \) by \(−2\), and adding it to equation \( (E1) \). We then perform the same steps as above and find the same result, \(0=0\).

When a system is dependent, we can find general expressions for the solutions. Adding equations \( (E1) \) and \( (E3) \), we have\[ \begin{array}{crccccccc} & (E1): & 2x & + & y & - & 3z & = & 0 \\[6pt] + & (E3): & x & - & y & + & z & = & 0 \\[6pt] \hline & & 3x & & & - & 2z & = & 0 \\[6pt] \end{array} \nonumber \]We then solve the resulting equation for \(z\).\[3x−2z=0 \implies z = \dfrac{3}{2} x \nonumber \]We back-substitute the expression for \(z\) into one of the equations and solve for \(y\).\[ \begin{array}{rccccccccl} & 2x & + & y & - & 3z & = & 0 & & \\[6pt] \implies & 2x & + & y & - & 3\left( \dfrac{3}{2} x \right) & = & 0 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 2x & + & y & - & \dfrac{9}{2} x & = & 0 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & -\dfrac{5}{2} x & + & y & & & = & 0 & \quad & \left( \text{combining like terms} \right) \\[6pt] \implies & & & y & & & = & \dfrac{5}{2} x & \quad & \left( \text{adding }\frac{5}{2}x \text{ to both sides} \right) \\[6pt] \end{array} \nonumber \]So the general solution is \(\left( x, \frac{5}{2}x, \frac{3}{2} x \right)\). In this solution, \(x\) can be any real number. The values of \(y\) and \(z\) are dependent on the value selected for \(x\).

As shown in Figure \( \PageIndex{ 5 } \), two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.

Figure \( \PageIndex{ 5 } \)
 

Q&A

Does the generic solution to a dependent system always have to be written in terms of \(x\)?

No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of \( x \) or, if needed, \(x\) and \(y\).

Checkpoint \( \PageIndex{ 6 } \)

Solve the following system.\[ \begin{cases} x & + & y & + & z & = & 7 \\ 3x & − & 2y & − & z & = & 4 \\ x & + & 6y & + & 5z & = & 24 \\ \end{cases} \nonumber \]