Lesson 4.4: Solve Systems of Equations with Three Variables

New Concept

Ready to move into 3D? This lesson extends your skills to systems with three variables. You'll learn to find the single point—an ordered triple $(x, y, z)$—where three planes intersect and apply this to solve real-world problems.

What’s next

First, you'll verify solutions with practice cards. Then, we'll dive into worked examples and challenge problems for solving entire systems.

Property

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.

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 $(x, y, z)$.

Examples

• Is $(2, -1, 3)$ a solution to the system $\begin{cases} x+y+z=4 \\ 2x-y-z=2 \\ 3x+2y+z=7 \end{cases}$? Yes, because substituting the values makes all three equations true: $2+(-1)+3=4$, $2(2)-(-1)-3=2$, and $3(2)+2(-1)+3=7$.

Property

To solve a system of linear equations with three variables**

• Step 1. Write the equations in standard form. If any coefficients are fractions, clear them.
• Step 2. Eliminate the same variable from two equations.
• Step 3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
• Step 4. The two new equations form a system of two equations with two variables. Solve this system.
• Step 5. Use the values of the two variables found in Step 4 to find the third variable.
• Step 6. Write the solution as an ordered triple.
• Step 7. Check that the ordered triple is a solution to all three original equations.

Examples

• Solve $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ x+2y-z=2 \end{cases}$. Adding (1) and (3) gives $2x+3y=8$. Adding (2) and (3) gives $3x+y=5$, or $y=5-3x$. Substituting into $2x+3y=8$ gives $2x+3(5-3x)=8$, so $x=1$. Then $y=2$ and $z=3$. The solution is $(1, 2, 3)$.

Property

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.

Examples

• Solve $\begin{cases} x-y+z=1 \\ -x+y-z=2 \\ 2x+y+z=3 \end{cases}$. Adding the first two equations immediately results in the false statement $0=3$. This system is inconsistent and has no solution.

• Solve $\begin{cases} x+y+z=2 \\ x+y+z=4 \\ x-y-z=0 \end{cases}$. Subtracting the first equation from the second gives the false statement $0=2$. The system is inconsistent.

Property

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.

Examples

• Solve $\begin{cases} x+y+z=4 \\ 2x+2y+2z=8 \\ x-y+z=2 \end{cases}$. The second equation is twice the first, leading to $0=0$. Using (1) and (3), add them to get $2x+2z=6$, so $x=3-z$. Substitute into (1): $(3-z)+y+z=4$, so $y=1$. The solution is $(3-z, 1, z)$.

• Solve $\begin{cases} x-2y+z=5 \\ 3x-6y+3z=15 \\ -2x+4y-2z=-10 \end{cases}$. All three equations are multiples of each other. This results in $0=0$. The solutions are all points on the plane $x-2y+z=5$. We can write $x=5+2y-z$.

Property

To solve applications, assign variables to the unknown quantities. Translate the problem's sentences into a system of three equations. Solve the system using elimination and state the answer in the context of the problem.

Examples

• A bakery sells cookies ($c$), brownies ($b$), and muffins ($m$). It sold 100 items for 240 dollars. Cookies cost 2 dollars, brownies 3 dollars, and muffins 2 dollars. It sold twice as many cookies as brownies. The system is $\begin{cases} c+b+m=100 \\ 2c+3b+2m=240 \\ c=2b \end{cases}$. The solution is 40 cookies, 20 brownies, and 40 muffins.

• An investor puts a total of 10,000 dollars into three funds: a safe fund ($x$), a moderate fund ($y$), and a high-risk fund ($z$). The total annual interest was 500 dollars. The funds yield 3%, 5%, and 8% respectively. The amount in the safe fund was equal to the sum of the other two. The system is $\begin{cases} x+y+z=10000 \\ 0.03x+0.05y+0.08z=500 \\ x=y+z \end{cases}$. The solution is 5000 dollars in the safe fund, 2000 dollars in moderate, and 3000 dollars in high-risk.