• 4.4 Solve Systems of Equations with Three Variables
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • 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

Be Prepared 4.10

Before you get started, take this readiness quiz.

Evaluate 5 x − 2 y + 3 z 5 x − 2 y + 3 z when x = −2 , x = −2 , y = −4 , y = −4 , and z = 3 . z = 3 . If you missed this problem, review Example 1.21 .

Be Prepared 4.11

Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 . If you missed this problem, review Example 2.6 .

Be Prepared 4.12

Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 . If you missed this problem, review Example 2.8 .

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 , a x + b y = c , a x + b y = c , is a line. Each point on the line, an ordered pair ( x , y ) , ( 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.

Similarly, for a linear equation with three variables a x + b y + c z = d , a x + b y + c z = d , every solution to the equation is an ordered triple, ( x , y , z ) ( 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

Every solution to the equation is an ordered triple, ( x , y , z ) ( 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.

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 ) ( 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 ) . ( 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 4.31

Determine whether the ordered triple is a solution to the system: { x − y + z = 2 2 x − y − z = −6 2 x + 2 y + z = −3 . { x − y + z = 2 2 x − y − z = −6 2 x + 2 y + z = −3 .

ⓐ ( −2 , −1 , 3 ) ( −2 , −1 , 3 ) ⓑ ( −4 , −3 , 4 ) ( −4 , −3 , 4 )

Try It 4.61

Determine whether the ordered triple is a solution to the system: { 3 x + y + z = 2 x + 2 y + z = −3 3 x + y + 2 z = 4 . { 3 x + y + z = 2 x + 2 y + z = −3 3 x + y + 2 z = 4 .

ⓐ ( 1 , −3 , 2 ) ( 1 , −3 , 2 ) ⓑ ( 4 , −1 , −5 ) ( 4 , −1 , −5 )

Try It 4.62

Determine whether the ordered triple is a solution to the system: { x − 3 y + z = −5 − 3 x − y − z = 1 2 x − 2 y + 3 z = 1 . { x − 3 y + z = −5 − 3 x − y − z = 1 2 x − 2 y + 3 z = 1 .

ⓐ ( 2 , −2 , 3 ) ( 2 , −2 , 3 ) ⓑ ( −2 , 2 , 3 ) ( −2 , 2 , 3 )

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 4.32

How to solve a system of equations with three variables by elimination.

Solve the system by elimination: { x − 2 y + z = 3 2 x + y + z = 4 3 x + 4 y + 3 z = −1 . { x − 2 y + z = 3 2 x + y + z = 4 3 x + 4 y + 3 z = −1 .

Try It 4.63

Solve the system by elimination: { 3 x + y − z = 2 2 x − 3 y − 2 z = 1 4 x − y − 3 z = 0 . { 3 x + y − z = 2 2 x − 3 y − 2 z = 1 4 x − y − 3 z = 0 .

Try It 4.64

Solve the system by elimination: { 4 x + y + z = −1 − 2 x − 2 y + z = 2 2 x + 3 y − z = 1 . { 4 x + y + z = −1 − 2 x − 2 y + z = 2 2 x + 3 y − z = 1 .

The steps are summarized here.

Solve a system of linear equations with three variables.

  • If any coefficients are fractions, clear them.
  • 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
  • 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.

Example 4.33

Solve: { 3 x − 4 z = 0 3 y + 2 z = −3 2 x + 3 y = −5 . { 3 x − 4 z = 0 3 y + 2 z = −3 2 x + 3 y = −5 .

We can eliminate z z from equations (1) and (2) by multiplying equation (2) by 2 and then adding the resulting equations.

Notice that equations (3) and (4) both have the variables x x and y y . We will solve this new system for x x and y y .

To solve for y , we substitute x = −4 x = −4 into equation (3).

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

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

We check that the solution makes all three equations true.

3 x − 4 z = 0 ( 1 ) 3 ( −4 ) − 4 ( −3 ) = ? 0 0 = 0 ✓ 3 y + 2 z = −3 ( 2 ) 3 ( 1 ) + 2 ( −3 ) = ? − 3 −3 = −3 ✓ 2 x + 3 y = −5 ( 3 ) 2 ( −4 ) + 3 ( 1 ) = ? − 5 −5 = −5 ✓ The solution is ( −4 , 1 , −3 ) . 3 x − 4 z = 0 ( 1 ) 3 ( −4 ) − 4 ( −3 ) = ? 0 0 = 0 ✓ 3 y + 2 z = −3 ( 2 ) 3 ( 1 ) + 2 ( −3 ) = ? − 3 −3 = −3 ✓ 2 x + 3 y = −5 ( 3 ) 2 ( −4 ) + 3 ( 1 ) = ? − 5 −5 = −5 ✓ The solution is ( −4 , 1 , −3 ) .

Try It 4.65

Solve: { 3 x − 4 z = −1 2 y + 3 z = 2 2 x + 3 y = 6 . { 3 x − 4 z = −1 2 y + 3 z = 2 2 x + 3 y = 6 .

Try It 4.66

Solve: { 4 x − 3 z = −5 3 y + 2 z = 7 3 x + 4 y = 6 . { 4 x − 3 z = −5 3 y + 2 z = 7 3 x + 4 y = 6 .

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 4.34

Solve the system of equations: { x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2 . { x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2 .

Use equation (1) and (2) to eliminate z .

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

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

There is no solution.

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

Try It 4.67

Solve the system of equations: { x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1 . { x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1 .

Try It 4.68

Solve the system of equations: { 2 x − 2 y + 3 z = 6 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 . { 2 x − 2 y + 3 z = 6 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 .

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 4.35

Solve the system of equations: { x + 2 y − z = 1 2 x + 7 y + 4 z = 11 x + 3 y + z = 4 . { x + 2 y − z = 1 2 x + 7 y + 4 z = 11 x + 3 y + z = 4 .

Use equation (1) and (3) to eliminate x .

Use equation (1) and (2) to eliminate x again.

Use equation (4) and (5) to eliminate y y .

The true statement 0 = 0 0 = 0 tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form ( x , y , z ) ( x , y , z ) where x = 5 z − 5 ; y = −2 z + 3 x = 5 z − 5 ; y = −2 z + 3 and z is any real number.

Try It 4.69

Solve the system by equations: { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 . { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 .

Try It 4.70

Solve the system by equations: { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 . { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 .

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 4.36

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?

Try It 4.71

The community college fine arts department sold three kinds of tickets to its latest dance presentation. The adult tickets sold for $20, the student tickets for $12 and the child tickets for $10.The fine arts department was thrilled to have sold 350 tickets and brought in $4,650 in one night. The number of child tickets sold is the same as the number of adult tickets sold. How many of each type did the department sell?

Try It 4.72

The community college soccer team sold three kinds of tickets to its latest game. The adult tickets sold for $10, the student tickets for $8 and the child tickets for $5. The soccer team was thrilled to have sold 600 tickets and brought in $4,900 for one game. The number of adult tickets is twice the number of child tickets. How many of each type did the soccer team sell?

Access this online resource for additional instruction and practice with solving a linear system in three variables with no or infinite solutions.

  • Solving a Linear System in Three Variables with No or Infinite Solutions
  • 3 Variable Application

Practice Makes Perfect

In the following exercises, determine whether the ordered triple is a solution to the system.

{ 2 x − 6 y + z = 3 3 x − 4 y − 3 z = 2 2 x + 3 y − 2 z = 3 { 2 x − 6 y + z = 3 3 x − 4 y − 3 z = 2 2 x + 3 y − 2 z = 3

ⓐ ( 3 , 1 , 3 ) ( 3 , 1 , 3 ) ⓑ ( 4 , 3 , 7 ) ( 4 , 3 , 7 )

{ − 3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1 { − 3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1

ⓐ ( −5 , −7 , 4 ) ( −5 , −7 , 4 ) ⓑ ( 5 , 7 , 4 ) ( 5 , 7 , 4 )

{ y − 10 z = −8 2 x − y = 2 x − 5 z = 3 { y − 10 z = −8 2 x − y = 2 x − 5 z = 3

ⓐ ( 7 , 12 , 2 ) ( 7 , 12 , 2 ) ⓑ ( 2 , 2 , 1 ) ( 2 , 2 , 1 )

{ x + 3 y − z = 15 y = 2 3 x − 2 x − 3 y + z = −2 { x + 3 y − z = 15 y = 2 3 x − 2 x − 3 y + z = −2

ⓐ ( −6 , 5 , 1 2 ) ( −6 , 5 , 1 2 ) ⓑ ( 5 , 4 3 , −3 ) ( 5 , 4 3 , −3 )

In the following exercises, solve the system of equations.

{ 5 x + 2 y + z = 5 − 3 x − y + 2 z = 6 2 x + 3 y − 3 z = 5 { 5 x + 2 y + z = 5 − 3 x − y + 2 z = 6 2 x + 3 y − 3 z = 5

{ 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1 { 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1

{ 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3

{ 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7 { 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7

{ 3 x − 5 y + 4 z = 5 5 x + 2 y + z = 0 2 x + 3 y − 2 z = 3 { 3 x − 5 y + 4 z = 5 5 x + 2 y + z = 0 2 x + 3 y − 2 z = 3

{ 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7 { 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7

{ 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1 { 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1

{ 11 x + 9 y + 2 z = −9 7 x + 5 y + 3 z = −7 4 x + 3 y + z = −3 { 11 x + 9 y + 2 z = −9 7 x + 5 y + 3 z = −7 4 x + 3 y + z = −3

{ 1 3 x − y − z = 1 x + 5 2 y + z = −2 2 x + 2 y + 1 2 z = −4 { 1 3 x − y − z = 1 x + 5 2 y + z = −2 2 x + 2 y + 1 2 z = −4

{ x + 1 2 y + 1 2 z = 0 1 5 x − 1 5 y + z = 0 1 3 x − 1 3 y + 2 z = −1 { x + 1 2 y + 1 2 z = 0 1 5 x − 1 5 y + z = 0 1 3 x − 1 3 y + 2 z = −1

{ x + 1 3 y − 2 z = −1 1 3 x + y + 1 2 z = 0 1 2 x + 1 3 y − 1 2 z = −1 { x + 1 3 y − 2 z = −1 1 3 x + y + 1 2 z = 0 1 2 x + 1 3 y − 1 2 z = −1

{ 1 3 x − y + 1 2 z = 4 2 3 x + 5 2 y − 4 z = 0 x − 1 2 y + 3 2 z = 2 { 1 3 x − y + 1 2 z = 4 2 3 x + 5 2 y − 4 z = 0 x − 1 2 y + 3 2 z = 2

{ x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3 { x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3

{ 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3 { 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3

{ 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1 { 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1

{ 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8 { 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8

{ 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6 { 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6

{ x − 2 y + 2 z = 1 − 2 x + y − z = 2 x − y + z = 5 { x − 2 y + 2 z = 1 − 2 x + y − z = 2 x − y + z = 5

{ 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20 { 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20

{ x + 4 y + z = −8 4 x − y + 3 z = 9 2 x + 7 y + z = 0 { x + 4 y + z = −8 4 x − y + 3 z = 9 2 x + 7 y + z = 0

{ x + 2 y + z = 4 x + y − 2 z = 3 − 2 x − 3 y + z = −7 { x + 2 y + z = 4 x + y − 2 z = 3 − 2 x − 3 y + z = −7

{ x + y − 2 z = 3 − 2 x − 3 y + z = −7 x + 2 y + z = 4 { x + y − 2 z = 3 − 2 x − 3 y + z = −7 x + 2 y + z = 4

{ x + y − 3 z = −1 y − z = 0 − x + 2 y = 1 { x + y − 3 z = −1 y − z = 0 − x + 2 y = 1

{ x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7 { x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7

In the following exercises, solve the given problem.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is twice the measure of the first angle. The third angle is twelve more than the second. Find the measures of the three angles.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is three times the measure of the first angle. The third angle is fifteen more than the second. Find the measures of the three angles.

After watching a major musical production at the theater, the patrons can purchase souvenirs. If a family purchases 4 t-shirts, the video, and 1 stuffed animal, their total is $135.

A couple buys 2 t-shirts, the video, and 3 stuffed animals for their nieces and spends $115. Another couple buys 2 t-shirts, the video, and 1 stuffed animal and their total is $85. What is the cost of each item?

The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of $65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of $140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 cans of popcorn for a total sales of $250. What is the cost of each item?

Writing Exercises

In your own words explain the steps to solve a system of linear equations with three variables by elimination.

How can you tell when a system of three linear equations with three variables has no solution? Infinitely many solutions?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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Systems of Equations with Three Variables Worksheets

Focusing on solving systems of equations with three variables, this assemblage of printable worksheets provides immense practice to high school students. Featured here are simultaneous equations to be solved using the substitution method, the elimination method, Cramer's rule, and involves reciprocal equations as well. Included here is a review section allowing students to use a method of their choice. Click on our free worksheets and kick-start practice!

Substitution Method

Substitution Method

Express one variable in terms of the other and substitute until one equation with one variable is left. Solve for the variable and perform back substitution to find the value of the other two, in this batch of pdf worksheets.

  • Download the set

Elimination Method

Elimination Method

Pair-up the linear equations to eliminate one variable and form two new equations of 2 variables each. Find the value of one variable by eliminating the other. Substitute and find the other two unknown variables in each system of equations.

Cramer's Rule

Cramer's Rule

Extract the coefficient, x, y and z matrices from the systems of equations. Find the determinant of each 3 x 3 matrix. Divide the determinant of the x, y and z matrices with the coefficient matrix to find the solution to each system of equations with 3 variables.

Solve using Any Method

Solve using Any Method

Each printable worksheet in this unit of solving systems of equations offers eight sets of equations. High school students use their discretion to choose from the substitution method, elimination method or the Cramer's Rule to find the solution to the systems of equations involving 3 variables.

Solving Reciprocal Equations

Solving Reciprocal Equations

Replace each fraction in the equation with a variable to convert the reciprocal equations into standard equations. Apply any method of your choice and determine the solution.

Related Worksheets

» Systems of Equations with Two Variables

» Rearranging Equations

» One-step Equation

» Two-Step Equation

» Multi-Step Equation

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Mathematics LibreTexts

3.4: Solving Linear Systems with Three Variables

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  • Page ID 6243

Learning Objectives

  • Check solutions to linear systems with three variables.
  • Solve linear systems with three variables by elimination.
  • Identify dependent and inconsistent systems.
  • Solve applications involving three unknowns.

Solutions to Linear Systems with Three Variables

Real-world applications are often modeled using more than one variable and more than one equation. In this section, we will study linear systems consisting of three linear equations each with three variables. For example,

\(\left\{ \begin{array} { l l } { 3 x + 2 y - z = - 7 } & { \color{Cerulean} { (1) } } \\ { 6 x - y + 3 z = - 4 } & { \color{Cerulean} { (2) } } \\ { x + 10 y - 2 z = 2 } & { \color{Cerulean} { (3) } } \end{array} \right.\)

A solution to such a linear system is an ordered triple 19 \((x, y, z)\) that solves all of the equations. In this case, \((−2, 1, 3)\) is the only solution. To check that an ordered triple is a solution, substitute in the corresponding \(x\)-, \(y\)-, and \(z\)-values and then simplify to see if you obtain a true statement from all three equations.

Because the ordered triple satisfies all three equations we conclude that it is indeed a solution.

Example \(\PageIndex{1}\):

Determine whether or not \((1, 4, \frac{4}{3}\) is a solution to the following linear system:

\(\left\{ \begin{aligned} 9 x + y - 6 z & = 5 \\ - 6 x - 3 y + 3 z & = - 14 \\ 3 x + 2 y - 7 z & = 15 \end{aligned} \right.\)

The point does not satisfy all of the equations and thus is not a solution.

An ordered triple such as \((2, 4, 5)\) can be graphed in three-dimensional space as follows:

926cca2c2942c813865579781eab4bdf.png

The ordered triple indicates position relative to the origin \((0, 0, 0)\), in this case, \(2\) units along the \(x\)-axis, \(4\) units parallel to the \(y\)-axis, and \(5\) units parallel to the \(z\)-axis. A linear equation with three variables 20 is in standard form if

\(ax+by+cz=d\)

where \(a, b, c\), and \(d\) are real numbers. For example, \(6x + y + 2z = 26\) is in standard form. Solving for \(z\), we obtain \(z = −3x − \frac{1}{2} y + 13\) and can consider both \(x\) and \(y\) to be the independent variables. When graphed in three-dimensional space, its graph will form a straight flat surface called a plane 21 .

cb50e8c9a978a4ca62a8c6d4a30d4c65.png

Therefore, the graph of a system of three linear equations and three unknowns will consist of three planes in space. If there is a simultaneous solution, the system is consistent and the solution corresponds to a point where the three planes intersect.

7459f0f0ab8b4fef9fb707a49ccbc0ae.png

Graphing planes in three-dimensional space is not within the scope of this textbook. However, it is always important to understand the geometric interpretation.

Exercise \(\PageIndex{1}\)

Determine whether or not \((3, −1, 2)\) a solution to the system:

\(\left\{ \begin{array} { l } { 2 x - 3 y - z = 7 } \\ { 3 x + 5 y - 3 z = - 2 } \\ { 4 x - y + 2 z = 17 } \end{array} \right.\)

Yes, it is a solution.

www.youtube.com/v/2UET4LzXoYg

Solve Linear Systems with Three Variables by Elimination

In this section, the elimination method is used to solve systems of three linear equations with three variables. The idea is to eliminate one of the variables and resolve the original system into a system of two linear equations, after which we can then solve as usual. The steps are outlined in the following example.

Example \(\PageIndex{2}\)

Solve: \(\left\{ \begin{array} { l l } { 3 x + 2 y - z = - 7 } & { \color{Cerulean}{(1)} } \\ { 6 x - y + 3 z = - 4 } & { \color{Cerulean} { (2) } } \\ { x + 10 y - 2 z = 2 } & { \color{Cerulean} { (3) } } \end{array} \right.\)

All three equations are in standard form. If this were not the case, it would be a best practice to rewrite the equations in standard form before beginning this process.

Step 1 : Choose any two of the equations and eliminate a variable. In this case, we can line up the variable \(z\) to eliminate if we group \(3\) times the first equation with the second equation.

516fe5445b196be3005983298be1a411.png

Next, add the equations together.

\(\begin{aligned} 9 x + 6 y \color{red}{- 3 z}&\color{black}{ =} 21 \\ \pm 6 x - y \color{red}{+ 3 z}&\color{black}{ =} - 4 \\ \hline \\ 15x + 5y &= -25 \color{OliveGreen}{✓} \end{aligned}\)

Step 2 : Choose any other two equations and eliminate the same variable. We can line up \(z\) to eliminate again if we group \(−2\) times the first equation with the third equation.

9ebe0a82e20ab5ef2e179afa669cdc14.png

And then add,

\(\begin{aligned} - 6 x - 4 y \color{red}{+ 2 z}&\color{black}{ =} 14 \\ \pm x + 10 y \color{red}{- 2 z} &\color{black}{=} 2 \\ \hline \\-5x + 6y& = 16 \color{OliveGreen}{✓}\end{aligned}\)

Step 3 : Solve the resulting system of two equations with two unknowns. Here we solve by elimination. Multiply the second equation by \(3\) to line up the variable \(x\) to eliminate.

479855f616ff4062df8e26f9d8ad2d39.png

\(\begin{aligned} \color{red}{15 x}\color{black}{ +} 5 y &= - 25 \\ \pm \color{red}{- 15 x}\color{black}{ +} 18 y &= 48 \\ \hline\\23y&=23\\y&=1 \end{aligned}\)

Step 4 : Back substitute and determine all of the coordinates. To find x use the following,

\(\begin{aligned} 15 x + 5 y & = - 25 \\ 15 x + 5 ( \color{OliveGreen}{1}\color{black}{ )} & = - 25 \\ 15 x & = - 30 \\ x & = - 2 \end{aligned}\)

Now choose one of the original equations to find \(z\),

\(\begin{aligned} 3 x + 2 y - z & = - 7 \quad\color{Cerulean}{(1)} \\ 3 ( \color{OliveGreen}{- 2}\color{Black}{ )} + 2 ( \color{OliveGreen}{1}\color{Black}{ )} - z & = - 7 \\ - 6 + 2 - z & = - 7 \\ - 4 - z & = - 7 \\ - z & = - 3 \\ z & = 3 \end{aligned}\)

Hence the solution, presented as an ordered triple \((x, y, z)\), is \((−2, 1, 3)\). This is the same system that we checked in the beginning of this section.

\((-2,1,3)\)

It does not matter which variable we initially choose to eliminate, as long as we eliminate it twice with two different sets of equations.

Example \(\PageIndex{3}\)

Solve: \(\left\{ \begin{array} { c } { - 6 x - 3 y + 3 z = - 14 } \\ { 9 x + y - 6 z = 5 } \\ { 3 x + 2 y - 7 z = 15 } \end{array} \right.\).

Because \(y\) has coefficient \(1\) in the second equation, choose to eliminate this variable. Use equations \(1\) and \(2\) to eliminate \(y\).

a930dc2e59fe26d4e9f63d71255af376.png

Next use the equations \(2\) and \(3\) to eliminate \(y\) again.

1851087164f84417d08d2fca715fe95b.png

This leaves a system of two equations with two variables \(x\) and \(z\),

\(\left\{ \begin{array} { l } { 21 x - 15 z = 1 } \\ { - 15 x + 5 z = 5 } \end{array} \right.\)

Multiply the second equation by \(3\) and eliminate the variable \(z\).

8f6730530de1044a573e39ba7e363c32.png

Now back substitute to find \(z\).

\(\begin{aligned} 21 x - 15 z & = 1 \\ 21 \left( \color{Cerulean}{- \frac { 2 } { 3} } \right) - 15 z & = 1 \\ - 14 - 15 z & = 1 \\ - 15 z & = 15 \\ z & = - 1 \end{aligned}\)

Finally, choose one of the original equations to find \(y\).

\(\begin{aligned} - 6 x - 3 y + 3 z & = - 14 \\ - 6 \left( \color{Cerulean}{- \frac { 2 } { 3} } \right) - 3 y + 3 ( \color{Cerulean}{- 1}\color{Black}{ )} & = - 14 \\ 4 - 3 y - 3 & = - 14 \\ 1 - 3 y & = - 14 \\ - 3 y & = - 15 \\ y & = 5 \end{aligned}\)

\(\left( - \frac { 2 } { 3 } , 5 , - 1 \right)\)

Example \(\PageIndex{4}\)

Solve: \(\left\{ \begin{array} { l } { 2 x + 6 y + 7 z = 4 } \\ { - 3 x - 4 y + 5 z = 12 } \\ { 5 x + 10 y - 3 z = - 13 } \end{array} \right.\).

In this example, there is no obvious choice of variable to eliminate. We choose to eliminate \(x\).

\(\begin{array} { l } { \color{Cerulean}{ (1) } } \\ { \color{Cerulean}{ (2) } } \end{array}\)\(\left \{ \begin{array} { l l } { 2 x + 6y + 7z = 4 } & { \stackrel{ \times3 } { \Rightarrow } } \\ { -3 x -4y +5z = 12 } & { \underset { \times 2 } { \Rightarrow }} \end{array} \right. \left\{ \begin{array} { l } { 6 x + 18 y + 21z = 12 } \\ { -6x -8y +10z = 24 } \end{array} \\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad10y + 31z=36 \color{OliveGreen}{✓} \right.\)

Next use equations \(2\) and \(3\) to eliminate \(x\) again.

\(\begin{array} { l } { \color{Cerulean}{ (2) } } \\ { \color{Cerulean}{ (3) } } \end{array}\)\(\left \{ \begin{array} { l l } { -3 x -4y + 5z = 12 } & { \stackrel{ \times5 } { \Rightarrow } } \\ { 5 x +10y -3z = -13 } & { \underset { \times 3 } { \Rightarrow }} \end{array} \right. \left\{ \begin{array} { l } { -15 x -20 y + 25z = 60 } \\ { 15x +30y -9z = -39 }\end{array} \\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad10y + 16z=21 \color{OliveGreen}{✓} \right.\)

This leaves a system of two equations with two variables \(y\) and \(z\),

\(\left\{ \begin{array} { l } { 10 y + 31 z = 36 } \\ { 10 y + 16 z = 21 } \end{array} \right.\)

Multiply the first equation by \(−1\) as a means to eliminate the variable \(y\).

95d63ad5a9baddc000a1cc8f24c2cb5f.png

Now back substitute to find \(y\).

\(\begin{aligned} 10 y + 31 z & = 36 \\ 10 y + 31 ( \color{OliveGreen}{1}\color{Black}{ )} & = 36 \\ 10 y + 31 & = 36 \\ 10 y & = 5 \\ y & = \frac { 5 } { 10 } \\ y & = \frac { 1 } { 2 } \end{aligned}\)

Choose any one of the original equations to find \(x\).

\(\begin{aligned} 2 x + 6 y + 7 z & = 4 \\ 2 x + 6 \left( \color{OliveGreen}{\frac { 1 } { 2 }} \right) + 7 ( \color{OliveGreen}{1}\color{Black}{ )} & = 4 \\ 2 x + 3 + 7 & = 4 \\ 2 x + 10 & = 4 \\ 2 x & = - 6 \\ x & = - 3 \end{aligned}\)

\(\left( - 3 , \frac { 1 } { 2 } , 1 \right)\)

Exercise \(\PageIndex{2}\)

Solve: \(\left\{ \begin{array} { l } { 2 x - 3 y - z = 7 } \\ { 3 x + 5 y - 3 z = - 2 } \\ { 4 x - y + 2 z = 17 } \end{array} \right.\)

\((3, -1, 2)\)

www.youtube.com/v/CjSv8D3g2Ic

Dependent and Inconsistent Systems

Just as with linear systems with two variables, not all linear systems with three variables have a single solution. Sometimes there are no simultaneous solutions.

Example \(\PageIndex{5}\):

Solve the system: \(\left\{ \begin{aligned} 4 x - y + 3 z & = 5 \\ 21 x - 4 y + 18 z & = 7 \\ - 9 x + y - 9 z & = - 8 \end{aligned} \right.\).

In this case we choose to eliminate the variable \(y\).

\(\begin{array} { l } { \color{Cerulean}{ (1) } } \\ { \color{Cerulean}{ (3) } } \end{array}\)\(\left \{ \begin{array} { l l } { 4 x -y + 3z = 5 } & \\ { -9 x +y -9z = -8 } & \end{array} \right. \\\quad-5x -6z=-3 \color{OliveGreen}{✓} \)

Next use equations \(2\) and \(3\) to eliminate \(y\) again.

5bb34b567d214f6483e9deac2397a57c.png

\(\left\{ \begin{array} { l } { - 5 x - 6 z = - 3 } \\ { - 15 x - 18 z = - 25 } \end{array} \right.\)

Multiply the first equation by \(-3\) and eliminate the variable \(z\).

740c553bef9233cde375b618839805b1.png

Adding the resulting equations together leads to a false statement, which indicates that the system is inconsistent. There is no simultaneous solution.

\(\varnothing\)

d550ca0de7c5e923812a078819bc153d.png

Just as with linear systems with two variables, some linear systems with three variables have infinitely many solutions. Such systems are called dependent systems.

Example \(\PageIndex{6}\)

Solve the system: \(\left\{ \begin{array} { c } { 7 x - 4 y + z = - 15 } \\ { 3 x + 2 y - z = - 5 } \\ { 5 x + 12 y - 5 z = - 5 } \end{array} \right.\).

Eliminate \(z\) by adding the first and second equations together.

\(\begin{array} { l } { \color{Cerulean}{ (1) } } \\ { \color{Cerulean}{ (3) } } \end{array}\)\(\left \{ \begin{array} { l l } { 7 x -4y + z = -15 } & \\ { 3 x +2y -z = -5 } & \end{array} \right. \\\quad10x -2y=-20 \color{OliveGreen}{✓} \)

Next use equations \(1\) and \(3\) to eliminate \(z\) again.

670a2b19d8ed509edab8ce48b9d5376b.png

This leaves a system of two equations with two variables \(x\) and \(y\),

\(\left\{ \begin{array} { l } { 10 x - 2 y = - 20 } \\ { 40 x - 8 y = - 80 } \end{array} \right.\)

Line up the variable \(y\) to eliminate by dividing the first equation by \(2\) and the second equation by \(−8\).

\(\left \{ \begin{array} { l l } { 10 x -2y = -20 } & { \stackrel{ \div 2} { \Longrightarrow } } \\ { 40 x -8y = -80 } & { \underset { \div (-8) } { \Longrightarrow }} \end{array} \right. \left\{ \begin{array} { l } { 5 x - y = -10 } \\ { -5x +y = 10 }\end{array} \\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 0=0 \:\:\color{Cerulean}{True} \right.\)

A true statement indicates that the system is dependent. To express the infinite number of solutions \((x, y,z)\) in terms of one variable, we solve for \(y\) and \(z\) both in terms of \(x\).

\(\begin{aligned} 10 x - 2 y & = - 20 \\ - 2 y & = - 10 x - 20 \\ \frac { - 2 y } { - 2 } & = \frac { - 10 x - 20 } { - 2 } \\ y & = 5 x + 10 \end{aligned}\)

Once we have \(y\) in terms of \(x\), we can solve for \(z\) in terms of \(x\) by back substituting into one of the original equations.

\(\begin{aligned} 7 x - 4 y + z & = - 15 \\ 7 x - 4 ( \color{OliveGreen}{5 x + 10}\color{black}{ )} + z & = - 15 \\ 7 x - 20 x - 40 + z & = - 15 \\ - 13 x - 40 + z & = - 15 \\ z & = 13 x + 25 \end{aligned}\)

\(( x , 5 x + 10,13 x + 25 )\)

A consistent system with infinitely many solutions is a dependent system. Given three planes, infinitely many simultaneous solutions can occur in a number of ways.

fba9f2933c40eda5a889e37a0e9c0abf.png

Exercise \(\PageIndex{3}\)

Solve: \(\left\{ \begin{aligned} 7 x + y - 2 z & = - 4 \\ - 21 x - 7 y + 8 z & = 4 \\ 7 x + 3 y - 3 z & = 0 \end{aligned} \right.\)

\(\left( x , \frac { 7 } { 3 } x + 4 , \frac { 14 } { 3 } x + 4 \right)\)

www.youtube.com/v/WgGAtNiJMlI

Applications Involving Three Unknowns

Many real-world applications involve more than two unknowns. When an application requires three variables, we look for relationships between the variables that allow us to write three equations.

Example \(\PageIndex{7}\)

A community theater sold \(63\) tickets to the afternoon performance for a total of \($444\). An adult ticket cost \($8\), a child ticket cost \($4\), and a senior ticket cost \($6\). If twice as many tickets were sold to adults as to children and seniors combined, how many of each ticket were sold?

Begin by identifying three variables.

Let \(x\) represent the number of adult tickets sold.

Let \(y\) represent the number of child tickets sold.

Let \(z\) represent the number of senior tickets sold.

The first equation comes from the statement that \(63\) tickets were sold.

\(\color{Cerulean}{(1)}\quad \color{black}{x}+y+z=63\)

The second equation comes from total ticket sales.

\(\color{Cerulean}{(2)}\quad \color{black}{8}x+4y+6z=444\)

The third equation comes from the statement that twice as many adult tickets were sold as child and senior tickets combined.

\(\begin{aligned} x & = 2 ( y + z ) \\ x & = 2 y + 2 z \\ \color{Cerulean}{( 3 )} \quad\color{black}{ x} - 2 y - 2 z & = 0 \end{aligned}\)

Therefore, the problem is modeled by the following linear system.

\(\left\{ \begin{array} { c } { x + y + z = 63 } \\ { 8 x + 4 y + 6 z = 444 } \\ { x - 2 y - 2 z = 0 } \end{array} \right.\)

Solving this system is left as an exercise. The solution is \((42, 9, 12)\).

The theater sold \(42\) adult tickets, \(9\) child tickets, and \(12\) senior tickets.

Key Takeaways

  • A simultaneous solution to a linear system with three equations and three variables is an ordered triple \((x, y, z)\) that satisfies all of the equations. If it does not solve each equation, then it is not a solution.
  • We can solve systems of three linear equations with three unknowns by elimination. Choose any two of the equations and eliminate a variable. Next choose any other two equations and eliminate the same variable. This will result in a system of two equations with two variables that can be solved by any method learned previously.
  • If the process of solving a system leads to a false statement, then the system is inconsistent and has no solution.
  • If the process of solving a system leads to a true statement, then the system is dependent and has infinitely many solutions.
  • To solve applications that require three variables, look for relationships between the variables that allow you to write three linear equations.

Exercise \(\PageIndex{4}\)

Determine whether or not the given ordered triple is a solution to the given system.

1. \((3, -2, -1)\);

\(\left\{ \begin{array} { c } { x + y - z = 2 } \\ { 2 x - 3 y + 2 z = 10 } \\ { x + 2 y + z = - 3 } \end{array} \right.\)

2. \((-8, -1, 5)\);

\(\left\{ \begin{array} { c } { x + 2 y - z = - 15 } \\ { 2 x - 6 y + 2 z = 0 } \\ { 3 x - 9 y + 4 z = 5 } \end{array} \right.\)

3. \((1, -9, 2)\);

\( \left\{ \begin{array} { c } { 8 x + y - z = - 3 } \\ { 7 x - 2 y - 3 z = 19 } \\ { x - y + 9 z = 28 } \end{array} \right.\)

4. \((-4, 1, -3)\);

\(\left\{ \begin{array} { l } { 3 x + 2 y - z = - 7 } \\ { x - 5 y + 2 z = 3 } \\ { 2 x + y + 3 z = - 16 } \end{array} \right.\)

5. \(\left( 6 , \frac { 2 } { 3 } , - \frac { 1 } { 2 } \right)\);

\(\left\{ \begin{array} { c } { x + 6 y - 4 z = 12 } \\ { - x + 3 y - 2 z = - 3 } \\ { x - 9 y + 8 z = - 4 } \end{array} \right.\)

6. \(\left( \frac { 1 } { 4 } , - 1 , - \frac { 3 } { 4 } \right)\);

\(\left\{ \begin{array} { r } { 2 x - y - 2 z = 3 } \\ { 4 x + 5 y - 8 z = 2 } \\ { x - 2 y - z = 3 } \end{array} \right.\)

7. \((3, -2, 1)\);

\(\left\{ \begin{array} { c } { 4 x - 5 y = 22 } \\ { 2 y - z = 8 } \\ { - 5 x + 2 z = - 13 } \end{array} \right.\)

8. \(\left( 1 , \frac { 5 } { 2 } , - \frac { 1 } { 2 } \right)\);

\(\left\{ \begin{aligned} 2 y - 6 z & = 8 \\ 3 x - 4 z & = 5 \\ 18 z & = - 9 \end{aligned} \right.\)

9. \(\left( \frac { 1 } { 2 } , - 2,6 \right)\);

  • \(\left\{ \begin{array} { c } { a - b + c = 9 } \\ { 4 a - 2 b + c = 14 } \\ { 2 a + b + \frac { 1 } { 2 } c = 3 } \end{array} \right.\)

10. \(( - 1,5 , - 7 )\);

  • \(\left\{ \begin{array} { l } { 3 a + b + \frac { 1 } { 3 } c = - \frac { 1 } { 3 } } \\ { 8 a + 2 b + \frac { 1 } { 2 } c = - \frac { 3 } { 2 } } \\ { 25 a + 5 b + c = - 7 } \end{array} \right.\)

Exercise \(\PageIndex{5}\)

  • \(\left\{ \begin{array} { l } { 2 x - 3 y + z = 4 } \\ { 5 x + 2 y + 2 z = 2 } \\ { x + 4 y - 3 z = 7 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 5 x - 2 y + z = - 9 } \\ { 2 x + y - 3 z = - 5 } \\ { 7 x + 3 y + 2 z = 6 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { x + 5 y - 2 z = 15 } \\ { 3 x - 7 y + 4 z = - 7 } \\ { 2 x + 4 y - 3 z = 21 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { x - 4 y + 2 z = 3 } \\ { 2 x + 3 y - 3 z = 9 } \\ { 3 x + 2 y + 4 z = - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 5 x + 4 y - 2 z = - 5 } \\ { 4 x - y + 3 z = 14 } \\ { 6 x + 3 y - 5 z = - 12 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 2 x + 3 y - 2 z = - 4 } \\ { 3 x + 5 y + 3 z = 17 } \\ { 2 x + y - 4 z = - 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { x + y - 4 z = 1 } \\ { 9 x - 3 y + 6 z = 2 } \\ { - 6 x + 2 y - 4 z = - 2 } \end{array} \right.\)
  • \(\left\{ \begin{aligned} 5 x - 8 y + z & = 5 \\ - 3 x + 5 y - z & = - 3 \\ - 11 x + 18 y - 3 z & = - 5 \end{aligned} \right.\)
  • \(\left\{ \begin{aligned} x - y + 2 z & = 3 \\ 2 x - y + 3 z & = 2 \\ - x - 3 y + 4 z & = 1 \end{aligned} \right.\)
  • \(\left\{ \begin{array} { c } { x + y + z = 8 } \\ { x - y + 4 z = - 7 } \\ { - x - y + 2 z = 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 4 x - y + 2 z = 3 } \\ { 6 x + 3 y - 4 z = - 1 } \\ { 3 x - 2 y + 3 z = 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { x - 4 y + 6 z = - 1 } \\ { 3 x + 8 y - 2 z = 2 } \\ { 5 x + 2 y - 3 z = - 5 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 4 y - z = 7 } \\ { 5 x - 8 y + 3 z = 11 } \\ { 2 x + 6 y + z = 9 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x + y - 4 z = 6 } \\ { 6 x - 5 y + 3 z = 1 } \\ { 9 x + 3 y - 4 z = 10 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 7 x - 6 y + z = 8 } \\ { - x + 2 y - z = 4 } \\ { x + 2 y - 2 z = 14 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { - 9 x + 3 y + z = 3 } \\ { 12 x - 4 y - z = 2 } \\ { - 6 x + 2 y + z = 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x - 5 y - 4 z = - 5 } \\ { 4 x - 6 y + 3 z = - 22 } \\ { 6 x + 8 y - 5 z = 20 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 7 x + 4 y - 2 z = 8 } \\ { 2 x + 2 y + 3 z = - 4 } \\ { 3 x - 6 y - 7 z = 8 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 9 x + 7 y + 4 z = 8 } \\ { 4 x - 5 y - 6 z = - 11 } \\ { - 5 x + 2 y + 3 z = 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 3 x + 7 y + 2 z = - 7 } \\ { 5 x + 4 y + 3 z = 5 } \\ { 2 x - 3 y + 5 z = - 4 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 4 x - 3 y = 1 } \\ { 2 y - 3 z = 2 } \\ { 3 x + 2 z = 3 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 5 y - 3 z = - 28 } \\ { 3 x + 2 y = 8 } \\ { 4 y - 7 z = - 27 } \end{array} \right.\)
  • \(\left\{ \begin{aligned} 2 x + 3 y + z & = 1 \\ 6 y + z & = 4 \\ 2 z & = - 4 \end{aligned} \right.\)
  • \(\left\{ \begin{aligned} x - 3 y - 2 z & = 5 \\ 2 y + 6 z & = - 1 \\ 4 z & = - 6 \end{aligned} \right.\)
  • \(\left\{ \begin{aligned} 2 x & = 10 \\ 6 x - 5 y & = 30 \\ 3 x - 4 y - 2 z & = 3 \end{aligned} \right.\)
  • \(\left\{ \begin{array} { c } { 2 x + 7 z = 2 } \\ { - 4 y = 6 } \\ { 8 y + 3 z = 0 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 5 x + 7 y + 2 z = 4 } \\ { 12 x + 16 y + 4 z = 15 } \\ { 10 x + 13 y + 3 z = 14 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 8 x + 12 y - 8 z = 5 } \\ { 2 x + 3 y - 2 z = 2 } \\ { 4 x - 2 y + 5 z = - 1 } \end{array} \right.\)
  • \(\left\{ \begin{array} { c } { 17 x - 4 y - 3 z = - 2 } \\ { 5 x + \frac { 1 } { 2 } y - 2 z = - \frac { 9 } { 2 } } \\ { 2 x + 5 y - 4 z = - 13 } \end{array} \right.\)
  • \(\left\{ \begin{aligned} 3 x - 5 y - \frac { 1 } { 2 } z & = \frac { 7 } { 2 } \\ x - y - \frac { 1 } { 2 } z & = - \frac { 1 } { 2 } \\ 3 x - 8 y + z & = 11 \end{aligned} \right.\)
  • \(\left\{ \begin{array} { l } { 4 a - 2 b + 3 c = 9 } \\ { 3 a + 3 b - 5 c = - 6 } \\ { 10 a - 6 b + 5 c = 13 } \end{array} \right.\)
  • \(\left\{ \begin{array} { l } { 6 a - 2 b + 5 c = - 2 } \\ { 4 a + 3 b - 3 c = - 1 } \\ { 3 a + 5 b + 6 c = 24 } \end{array} \right.\)

1. \((2, -1, -3)\)

3. \((4, 1, -3)\)

5. \((1, -1, 3)\)

7. \(\varnothing\)

9. \((5, -10, -6)\)

11. \(\left( \frac { 1 } { 2 } , - 2 , - \frac { 1 } { 2 } \right)\)

13. \(\left( 3 , \frac { 1 } { 2 } , 0 \right)\)

15. \(\left( x , \frac { 3 } { 2 } x - 3,2 x - 10 \right)\)

17. \((1, -2, 6)\)

19. \((-1, 2, -2)\)

21. \((1, -3, 5)\)

23. \((1, 1, 0)\)

25. \((0, 1, -2)\)

27. \((5, 0, 6)\)

29. \(\varnothing\)

31. \(( x , 2 x - 1,3 x + 2 )\)

33. \((1, 2, 3)\)

Exercise \(\PageIndex{6}\)

Set up a system of equations and use it to solve the following.

  • The sum of three integers is \(38\). Two less than \(4\) times the smaller integer is equal to the sum of the others. The sum of the smaller and larger integer is equal to \(2\) more than twice that of the other. Find the integers.
  • The sum of three integers is \(40\). Three times the smaller integer is equal to the sum of the others. Twice the larger is equal to \(8\) more than the sum of the others. Find the integers.
  • The sum of the angles \(A, B\), and \(C\) of a triangle is \(180°\). The larger angle \(C\) is equal to twice the sum of the other two. Four times the smallest angle \(A\) is equal to the difference of angle \(C\) and \(B\). Find the angles.
  • The sum of the angles \(A, B\), and \(C\) of a triangle is \(180°\). Angle \(C\) is equal to the sum of the other two angles. Five times angle \(A\) is equal to the sum of angle \(C\) and \(B\). Find the angles.
  • A total of \($12,000\) was invested in three interest earning accounts. The interest rates were \(2\)%, \(4\)%, and \(5\)%. If the total simple interest for one year was \($400\) and the amount invested at \(2\)% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account?
  • Joe invested his \($6,000\) bonus in three accounts earning \(4 \frac{1}{2}\)% interest. He invested twice as much in the account earning \(4 \frac{1}{2}\)% as he did in the other two accounts combined. If the total simple interest for the year was \($234\), how much did Joe invest in each account?
  • A jar contains nickels, dimes, and quarters. There are \(105\) coins with a total value of \($8.40\). If there are \(3\) more than twice as many dimes as quarters, find how many of each coin are in the jar.
  • A billfold holds one-dollar, five-dollar, and ten-dollar bills and has a value of \($210\). There are \(50\) bills total where the number of one-dollar bills is one less than twice the number of five-dollar bills. How many of each bill are there?
  • A nurse wishes to prepare a \(15\)-ounce topical antiseptic solution containing \(3\)% hydrogen peroxide. To obtain this mixture, purified water is to be added to the existing \(1.5\)% and \(10\)% hydrogen peroxide products. If only \(3\) ounces of the \(10\)% hydrogen peroxide solution is available, how much of the \(1.5\)% hydrogen peroxide solution and water is needed?
  • A chemist needs to produce a \(32\)-ounce solution consisting of \(8 \frac{3}{4}\)% acid. He has three concentrates with \(5\)%, \(10\)%, and \(40\)% acid. If he is to use twice as much of the \(5\)% acid solution as the \(10\)% solution, then how many ounces of the \(40\)% solution will he need?
  • A community theater sold \(128\) tickets to the evening performance for a total of \($1,132\). An adult ticket cost \($10\), a child ticket cost \($5\), and a senior ticket cost \($6\). If three times as many tickets were sold to adults as to children and seniors combined, how many of each ticket were sold?
  • James sold \(82\) items at the swap meet for a total of \($504\). He sold packages of socks for \($6\), printed t-shirts for \($12\), and hats for \($5\). If he sold \(5\) times as many hats as he did t-shirts, how many of each item did he sell?
  • A parabola passes through three points \((−1, 7), (1, −1)\) and \((2, −2)\). Use these points and \(y = a x ^ { 2 } + b x + c\) to construct a system of three linear equations in terms of \(a, b\), and \(c\) and then solve the system.
  • A parabola passes through three points \((−2, 11), (−1, 4)\) and \((1, 2)\). Use these points and \(y = a x ^ { 2 } + b x + c\) to construct a system of three linear equations in terms of \(a, b\), and \(c\) and solve it.

1. \(8, 12, 18\)

3. \(A = 20 ^ { \circ } , B = 40 ^ { \circ } , \text { and } C = 120 ^ { \circ }\)

5. The amount invested at \(2\)% was \($6,000\), the amount invested at \(4\)% was \($2,000\), and the amount invested at \(5\)% was \($4,000\).

7. \(72\) nickels, \(23\) dimes, and \(10\) quarters

9. \(10\) ounces of the \(1.5\)% hydrogen peroxide solution and \(2\) ounces of water

11. \(96\) adult tickets, \(20\) child tickets, and \(12\) senior tickets were sold.

13. \(a = 1, b = −4\), and \(c = 2\)

Exercise \(\PageIndex{7}\)

  • On a note card, write down the steps for solving a system of three linear equations with three variables using elimination. Use your notes to explain to a friend how to solve one of the exercises in this section.
  • Research and discuss curve fitting. Why is curve fitting an important topic?

1. Answer may vary

19 Triples \((x, y, z)\) that identify position relative to the origin in three-dimensional space.

20 An equation that can be written in the standard form \(ax + by + cz = d\) where \(a, b, c\), and \(d\) are real numbers.

21 Any flat two-dimensional surface.

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Chapter 5: Systems of Equations

5.4 Solving for Three Variables

When given three variables, you are given the equation for a plane or a flat surface similar to a sheet of paper. Some of the possible solutions to the intersections of these equations can be visualized below.

Visualizations of different planes

In solving systems of equations with three variables, use the strategies that are used to solve systems of two equations. One recommended method is to eliminate one variable at the onset, thus turning the set of three equations with three unknowns into two equations with two unknowns. The standard method to work with three equations or more is to use subtraction and/or addition.

Example 5.4.1

Find the intersection or the solution to the following system of equations: [latex]3x+2y-z=-1, -2x-2y+3z=5,[/latex] and [latex]5x+2y-z=3.[/latex]

As we did with a set of two equations, first line up the equations to choose the variable that we wish to eliminate:

[latex]\left\{ \begin{array}{rrrrrrr} 3x&+&2y&-&z&=&-1 \\ -2x&-&2y&+&3z&=&5 \\ 5x&+&2y&-&z&=&3 \end{array} \right.[/latex]

For these equations, it looks easiest to eliminate the [latex]y[/latex]-variable. To do this, add the first and second equations together and then add the second and third equations together:

[latex]\begin{array}{rr} \begin{array}{rrrrrrrr} &3x&+&2y&-&z&=&-1 \\ +&-2x&-&2y&+&3z&=&5 \\ \hline &&&x&+&2z&=&4 \end{array} &\hspace{0.25in} \begin{array}{rrrrrrrr} &-2x&-&2y&+&3z&=&5 \\ +&5x&+&2y&-&z&=&3 \\ \hline &&&3x&+&2z&=&8 \end{array} \end{array}[/latex]

Now, you are left with [latex]x+2z=4[/latex] and [latex]3x + 2z = 8.[/latex] We now solve these as done previously with a set of two equations:

[latex]\left\{ \begin{array}{rrrrr} x&+&2z&=&4 \\ 3x&+&2z&=&8 \end{array}\right.[/latex]

Multiply either the top or the bottom equation by −1 to eliminate the [latex]z[/latex]-variable.

[latex]\begin{array}{rrrrrrr} &(x&+&2z&=&4)&(-1) \\ &3x&+&2z&=&8& \\ \\ &-x&-&2z&=&-4& \\ +&3x&+&2z&=&8& \\ \hline &&&\dfrac{2x}{2}&=&\dfrac{4}{2}& \\ \\ &&&x&=&2& \end{array}[/latex]

Next, find [latex]z[/latex] using one of [latex]x+2z=4[/latex] or [latex]3x+2z=8[/latex] and the solution [latex]x= 2.[/latex] [latex]x+2z=4[/latex] looks to be the easiest to work with.

[latex]\begin{array}{rrrrr} x&+&2z&=&4 \\ 2&+&2z&=&4 \\ -2&&&&-2 \\ \hline &&\dfrac{2z}{2}&=&\dfrac{2}{2} \\ \\ &&z&=&1 \end{array}[/latex]

Finally,  find [latex]y[/latex] using one of the original three equations:

[latex]\begin{array}{rrrrrrr} 3x&+&2y&-&z&=&-1 \\ 3(2)&+&2y&-&(1)&=&-1 \\ 6&+&2y&-&1&=&-1 \\ &&5&+&2y&=&-1 \\ &&-5&&&=&-5 \\ \hline &&&&\dfrac{2y}{2}&=&\dfrac{-6}{2} \\ \\ &&&&y&=&-3 \end{array}[/latex]

These planes intersect at the point [latex]x = 2,[/latex] [latex]y = -3,[/latex] and [latex]z = 1[/latex], or the coordinate [latex](2, -3, 1).[/latex]

Sometimes, you are given a set of three equations with missing variables. These systems of equations require slightly more thought to solve than the previous problems.

Example 5.4.2

Find the intersection or the solution to the following system of equations: [latex]x+2y-z=0, 3x-2y=-2,[/latex] and [latex]y+z=3.[/latex]

First, line up the equations to choose the variable that we wish to eliminate:

[latex]\left\{ \begin{array}{rrrrrrr} x&+&2y&-&z&=&0 \\ 3x&-&2y&&&=&-2 \\ &&y&+&z&=&3 \end{array}\right.[/latex]

In this example, adding the first and last equations eliminates the variable [latex]z,[/latex] without having to modify any of the equations:

[latex]\begin{array}{rrrrrrrr} &x&+&2y&-&z&=&0 \\ +&&&y&+&z&=&3 \\ \hline &&&x&+&3y&=&3 \\ \end{array}[/latex]

Now, there are two equations left:

[latex]\left\{ \begin{array}{rrrrr} 3x&-&2y&=&-2 \\ x&+&3y&=&3 \end{array}\right.[/latex]

First multiply the bottom equation by −3, then add it to the top equation, to eliminate the variable [latex]x[/latex]:

[latex]\begin{array}{rrrrrrr} &(x&+&3y&=&3)&(-3) \\ \\ &3x&-&2y&=&-2& \\ +&-3x&-&9y&=&-9& \\ \hline &&&-11y&=&-11& \\ &&&y&=&1& \\ \end{array}[/latex]

Now choose one of the two remaining equations, [latex]3x-2y=-2[/latex] or [latex]x+3y=3,[/latex] to find the variable [latex]x.[/latex] Choosing [latex]x+3y= 3,[/latex] leaves:

[latex]\begin{array}{rrrrr} x&+&3(1)&=&3 \\ x&+&3&=&3 \\ &-&3&=&-3 \\ \hline &&x&=&0 \end{array}[/latex]

Finally, to find the third variable, use one of the original three equations: [latex]x+2y-z=0, 3x-2y=-2,[/latex]  or [latex]y+z=3.[/latex] Choosing [latex]y + z = 3,[/latex] gives:

[latex]\begin{array}{rrrrr} (1)&+&z&=&3 \\ &&z&=&2 \end{array}[/latex]

These planes intersect at the point [latex]x = 0, y = 1,[/latex] and [latex]z = 2,[/latex] or the coordinate [latex](0, 1, 2).[/latex]

Solve each of the following systems of equations.

  • [latex]\left\{ \begin{array}{rrrrrrr} a&-&b&+&2c&=&2 \\ 2a&+&b&-&c&=&2 \\ a&+&b&+&c&=&3 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} 2a&+&3b&-&c&=&12 \\ 3a&+&4b&+&c&=&19 \\ a&-&2b&+&c&=&-3 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} 3x&+&y&-&z&=&7 \\ x&+&3y&-&z&=&5 \\ x&+&y&+&2z&=&3 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&y&+&z&=&4 \\ x&+&2y&+&3z&=&10 \\ x&-&y&+&4z&=&20 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&2y&-&z&=&0 \\ 2x&-&y&+&z&=&15 \\ 3x&-&2y&-&4z&=&-5 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&-&y&+&2z&=&-3 \\ x&+&2y&+&3z&=&4 \\ 2x&+&y&+&z&=&-3 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&y&+&z&=&6 \\ 2x&-&y&-&z&=&-3 \\ x&-&2y&+&3z&=&6 \end{array}\right.[/latex]
  • [latex]\text{tricky:} \left\{ \begin{array}{rrrrrrr} x&+&y&-&z&=&0 \\ x&+&2y&-&4z&=&0 \\ 2x&+&y&+&z&=&0 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&y&+&z&=&2 \\ 2x&-&y&+&3z&=&9 \\ &&y&-&z&=&-3 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} 6x&-&y&-&2z&=&-1 \\ 4x&&&+&z&=&3 \\ -2x&+&3y&&&=&5 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} &&y&+&z&=&5 \\ 2x&-&3y&+&z&=&-1 \\ x&&&-&z&=&-2 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} 3x&+&4y&-&z&=&11 \\ &&y&+&2z&=&-4 \\ -2x&+&y&&&=&-6 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&6y&+&3z&=&30 \\ 2x&&&+&2z&=&4 \\ &&-2y&+&z&=&-6 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&-&y&+&2z&=&0 \\ x&+&2y&&&=&1 \\ 2x&&&+&z&=&4 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&y&+&z&=&4 \\ &&-y&-&z&=&-4 \\ x&-&2y&&&=&0 \end{array}\right.[/latex]
  • [latex]\left\{ \begin{array}{rrrrrrr} x&+&y&-&z&=&2 \\ &&2y&-&4z&=&-4 \\ 2x&&&+&z&=&6 \end{array}\right.[/latex]

Answer Key 5.4

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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College Algebra : Linear Systems with Three Variables

Study concepts, example questions & explanations for college algebra, all college algebra resources, example questions, example question #1 : linear systems with three variables.

Solve this system of equations.

solving three variable equations practice problems

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

solving three variable equations practice problems

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

solving three variable equations practice problems

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

solving three variable equations practice problems

We can use this z value to find y

solving three variable equations practice problems

So the solution set is x = 1, y = 2, and z = –5/3.

Example Question #9 : Solving Equations

What is a solution to this system of equations:

solving three variable equations practice problems

Step 1:  Multiply first equation by  − 2  and add the result to the second equation. The result is:

solving three variable equations practice problems

Step 2:  Multiply first equation by  − 3  and add the result to the third equation. The result is:

solving three variable equations practice problems

Step 3:  Multiply second equation by  − 2 3  and add the result to the third equation. The result is:

solving three variable equations practice problems

Step 6:  solve for  x  by substituting  y = 2  and  z = 1  into the first equation.

solving three variable equations practice problems

Solve the following system of equations for a, b, and c.

solving three variable equations practice problems

To solve the equations we want to try to cancel out some of the letters to reduce the equations. We can do this by adding different equations to each other. If we add:

solving three variable equations practice problems

we get a new equation, 

solving three variable equations practice problems

We can also add 

solving three variable equations practice problems

So we can rearrange the new equations to get

solving three variable equations practice problems

and substitute these into the original second line equation:

solving three variable equations practice problems

Then we can solve for b and c:

solving three variable equations practice problems

Solve the following system of equations:

solving three variable equations practice problems

Add the first and third equations together, to eliminate z:

solving three variable equations practice problems

Add the first and second equations together to eliminate y:

solving three variable equations practice problems

Make sure the values for x,y, and z satisfy all of the equations:

solving three variable equations practice problems

Add the first two equations to eliminate the y and z-variable.

solving three variable equations practice problems

Using the value of x, with the first and third equations, we will need to eliminate the z-variable to solve for y.

solving three variable equations practice problems

Multiply the top equation by three.

solving three variable equations practice problems

Subtract both equations.

solving three variable equations practice problems

Divide both sides by two.

solving three variable equations practice problems

Example Question #4 : Linear Systems With Three Variables

Consider the system of linear equations

solving three variable equations practice problems

Describe the system.

The system is consistent and independent.

The system is inconsistent.

The system is coincident dependent.

The system is linearly dependent.

The given system has as many variables as equations, which makes it possible to have exactly one solution.

One way to identify the solution set is to use Gauss-Jordan elimination on the augmented coefficient matrix

solving three variable equations practice problems

Perform operations on the rows, with the object of rendering this matrix in reduced row-echelon form.

First, a 1 is wanted in Row 1, Column 1. This is already the case, so 0's are wanted elsewhere in Column 1. Do this using the row operation:

solving three variable equations practice problems

Next, a 1 is wanted in Row 2, Column 2. This is already the case, so 0's are wanted elsewhere in Column 2. Do this using the row operations:

solving three variable equations practice problems

Next, a 1 is wanted in Row 3, Column 3.Perform the operation

solving three variable equations practice problems

Now, 0's are wanted in the rest of Column 3:

solving three variable equations practice problems

Therefore, the solution to this system is

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How to solve systems of 3 variable equations

Using Elimination

Types of solutions for systems of planes (3 variable equations)

What is a solution of system of equations with 3 variables.

Solution icon

Solution for system of lines

Just as the solution system of lines is where those lines meet, a solution for a system of 3 variable equations (planes), is again, just where these planes meet.

types of solutions

read more here

Why 3 planes?

If you want to solve a linear equation with 2 variables, you need 2 equations.

You can's solve $$ x + y = 1$$ , right? That's because you need equations to solve for 2 variables.

Similarly, if you have an equation with 3 variables, ( graphically represented by 3 planes), you're going to need 3 equations to solve it.

Two important terms

Means that there is at least 1 intersection (solutions).

Means that there are no intersections (solutions).

Ok, so how do we find that point of intersection?

Before attempting to solve systems of three variable equations using elimination, you should probably be comfortable solving 2 variable systems of linear equations using elimination .

Video Tutorial on using Elimination

Example of how to solve a system of three variable equations using elimination.

Steps to solve system of 3 equations

Practice Problems

Use elimination to solve the following system of three variable equations.

  • A) 4x + 2y – 2z = 10
  • B) 2x + 8y + 4z = 32
  • C) 30x + 12y – 4z = 24

steps to solve 3 variable system by elimination

  • A) x - y + z = -1
  • B) x + y + z = 3
  • C) 4x + 2y + z = 8
  • A) 2x + 2y + 2z = -4

Although you can indeed solve 3 variable systems using elimination and substitution as shown on this page, you may have noticed that this method is quite tedious. The most efficient method is to use matrices or, of course, you can use this online system of equations solver . ( all of our pictures on this topic )

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

Popular pages @ mathwarehouse.com.

Surface area of a Cylinder

Course Enrichment Unit

Solve systems of three equations in three variables, learning outcomes.

  • Determine whether an ordered triple is a solution to a system of three equations.
  • Use back substitution to find a solution to a system of three equations.
  • Write the equations for a system given a scenario, and solve.

In order to solve systems of equations in three variables, known as three-by-three systems, the primary goal is to eliminate one variable at a time to achieve back-substitution. A solution to a system of three equations in three variables [latex]\left(x,y,z\right),\text{}[/latex] is called an ordered triple .

To find a solution, we can perform the following operations:

  • Interchange the order of any two equations.
  • Multiply both sides of an equation by a nonzero constant.
  • Add a nonzero multiple of one equation to another equation.

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.

A General Note: Number of Possible Solutions

The planes illustrate possible solution scenarios for three-by-three systems.

  • Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ordered triple [latex]\left\{\left(x,y,z\right)\right\}[/latex]. 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 [latex]0=0[/latex]. 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 [latex]3=0[/latex]. Graphically, a system with no solution is represented by three planes with no point in common.

solving three variable equations practice problems

(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.

solving three variable equations practice problems

Example: Determining Whether an Ordered Triple Is a Solution to a System

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]

We will check each equation by substituting in the values of the ordered triple for [latex]x,y[/latex], and [latex]z[/latex].

[latex]\begin{align} x+y+z=2\\ \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \text{True}\end{align}\hspace{5mm}[/latex] [latex]\hspace{5mm}\begin{align} 6x - 4y+5z=31\\ 6\left(3\right)-4\left(-2\right)+5\left(1\right)=31\\ 18+8+5=31\\ \text{True}\end{align}\hspace{5mm}[/latex] [latex]\hspace{5mm}\begin{align}5x+2y+2z=13\\ 5\left(3\right)+2\left(-2\right)+2\left(1\right)=13\\ 15 - 4+2=13\\ \text{True}\end{align}[/latex]

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.

  • Pick any pair of equations and solve for one variable.
  • Pick another pair of equations and solve for the same variable.
  • You have created a system of two equations in two unknowns. Solve the resulting two-by-two system.
  • Back-substitute known variables into any one of the original equations and solve for the missing variable.

tip for success

Work through the following examples on paper in the order given. They slowly build up the technique of solving a three-by-three system in stages. Then you will have an opportunity to practice the 4-step process given in the How To box above.

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 creates more steps in the solution process.

We’ll take the steps slowly in the following few examples. First, we’ll look just at the last step in the process: back-substitution. Then, we’ll look at an example that requires the addition (elimination) method to reach the first solution. Then we’ll see some video examples that illustrate some of the different kinds of situations you may encounter when solving three-by-three systems. Finally, you’ll have the opportunity to practice applying the complete process.

In the example that follows, we will solve the system by using back-substitution.

Solve the given system.

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

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]\begin{array}{l}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]

Now we have two of our solutions, and we can substitute them both into the first equation to solve for [latex]x[/latex].

[latex]\begin{array}{l}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]

Now we have our ordered triple; remember to place each variable solution in order.

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

Analysis of the Solution: Each of the lines in the system above 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.

In the next example we’ll need to use the addition method (elimination) to find our first solution.

Find a solution to the following system:

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).

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 is ok for variables to equal [latex]0[/latex].

Now we can substitute [latex]z = 6[/latex] and [latex]y = 0[/latex] back into the first equation.

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

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

Watch the following videos for more examples of the algebra you may encounter when solving systems with three variables.

Now, try the example and problems that follow to see if the process is becoming familiar to you. Solving three-by-three systems involves both creativity and careful, well-organized work. It will take some practice before it begins to feel natural.

Example: Solving a System of Three Equations in Three Variables by Elimination

[latex]\begin{align}x - 2y+3z=9& &\text{(1)} \\ -x+3y-z=-6& &\text{(2)} \\ 2x - 5y+5z=17& &\text{(3)} \end{align}[/latex]

There will always be several choices as to where to begin, but the most obvious first step here is to eliminate [latex]x[/latex] by adding equations (1) and (2).

[latex]\begin{align}x - 2y+3z&=9\\ -x+3y-z&=-6 \\ \hline y+2z&=3 \end{align}[/latex][latex]\hspace{5mm}\begin{gathered}\text{(1})\\ \text{(2)}\\ \text{(4)}\end{gathered}[/latex]

The second step is multiplying equation (1) by [latex]-2[/latex] and adding the result to equation (3). These two steps will eliminate the variable [latex]x[/latex].

[latex]\begin{align}−2x+4y−6z&=−18 \\ 2x−5y+5z&=17 \\ \hline −y−z&=−1\end{align}[/latex][latex]\hspace{5mm}\begin{align}&\text{(2) multiplied by }−2\\&\left(3\right)\\&(5)\end{align}[/latex]

In equations (4) and (5), we have created a new two-by-two system. We can solve for [latex]z[/latex] by adding the two equations.

[latex]\begin{align}y+2z&=3 \\ -y-z&=-1 \\ \hline z&=2 \end{align}[/latex][latex]\hspace{5mm}\begin{align}(4)\\(5)\\(6)\end{align}[/latex]

Choosing one equation from each new system, we obtain the upper triangular form:

[latex]\begin{align}x - 2y+3z&=9 && \left(1\right) \\ y+2z&=3 && \left(4\right) \\ z&=2 && \left(6\right) \end{align}[/latex]

Next, we back-substitute [latex]z=2[/latex] into equation (4) and solve for [latex]y[/latex].

[latex]\begin{align}y+2\left(2\right)&=3 \\ y+4&=3 \\ y&=-1 \end{align}[/latex]

Finally, we can back-substitute [latex]z=2[/latex] and [latex]y=-1[/latex] into equation (1). This will yield the solution for [latex]x[/latex].

[latex]\begin{align} x - 2\left(-1\right)+3\left(2\right)&=9\\ x+2+6&=9\\ x&=1\end{align}[/latex]

The solution is the ordered triple [latex]\left(1,-1,2\right)[/latex].

Three planes intersect at 1, negative 1, 2. The light blue plane's equation is x=1. The dark blue plane's equation is z=2. The orange plane's equation is y=negative 2.

Solve the system of equations in three variables.

[latex]\begin{array}{l}2x+y - 2z=-1\hfill \\ 3x - 3y-z=5\hfill \\ x - 2y+3z=6\hfill \end{array}[/latex]

[latex]\left(1,-1,1\right)[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=23765&theme=oea&iframe_resize_id=mom1

In the following video, you will see a visual representation of the three possible outcomes for solutions to a system of equations in three variables. There is also a worked example of solving a system using elimination.

Applications of Systems of Three Equations in Three Variables

Now we are ready to handle the problem we encountered as we began this section by using what we know about linear equations to translate the situation into a system of three equations. Then, we’ll use our new understanding of three-by-three systems to find the solution.

Tip for success

Applications of three-by-three systems are complicated. Work through each of the examples below perhaps more than once or twice. Don’t be discouraged if you don’t understand the process right away. It will take time and practice to become familiar with it.

Example: 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?

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:

[latex]\begin{align}&x=\text{amount invested in money-market fund} \\ &y=\text{amount invested in municipal bonds} \\ z&=\text{amount invested in mutual funds} \end{align}[/latex]

The first equation indicates that the sum of the three principal amounts is $12,000.

[latex]x+y+z=12{,}000[/latex]

We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.

[latex]z=y+4{,}000[/latex]

The third equation shows that the total amount of interest earned from each fund equals $670.

[latex]0.03x+0.04y+0.07z=670[/latex]

Then, we write the three equations as a system.

[latex]\begin{align}x+y+z=12{,}000 \\ -y+z=4{,}000 \\ 0.03x+0.04y+0.07z=670 \end{align}[/latex]

To make the calculations simpler, we can multiply the third equation by 100. Thus,

[latex]\begin{align}x+y+z=12{,}000 \hspace{5mm} \left(1\right) \\ -y+z=4{,}000 \hspace{5mm} \left(2\right) \\ 3x+4y+7z=67{,}000 \hspace{5mm} \left(3\right) \end{align}[/latex]

Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

[latex]\begin{align}x+y+z=12{,}000\hfill \\ 3x+4y +7z=67{,}000 \\ -y+z=4{,}000 \end{align}[/latex]

Step 2. Multiply equation (1) by [latex]-3[/latex] and add to equation (2). Write the result as row 2.

[latex]\begin{align}x+y+z=12{,}000 \\ y+4z=31{,}000 \\ -y+z=4{,}000 \end{align}[/latex]

Step 3. Add equation (2) to equation (3) and write the result as equation (3).

[latex]\begin{align}x+y+z=12{,}000 \\ y+4z=31{,}000 \\ 5z=35{,}000 \end{align}[/latex]

Step 4. Solve for [latex]z[/latex] in equation (3). Back-substitute that value in equation (2) and solve for [latex]y[/latex]. Then, back-substitute the values for [latex]z[/latex] and [latex]y[/latex] into equation (1) and solve for [latex]x[/latex].

[latex]\begin{align}&5z=35{,}000 \\ &z=7{,}000 \\ \\ &y+4\left(7{,}000\right)=31{,}000 \\ &y=3{,}000 \\ \\ &x+3{,}000+7{,}000=12{,}000 \\ &x=2{,}000 \end{align}[/latex]

John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.

https://ohm.lumenlearning.com/multiembedq.php?id=19353&theme=oea&iframe_resize_id=mom10

Systems of three equations in three variables apply to other types of real-world situations as well.

In this example, we will write three equations that model sales at an art fair to learn how many prints should be sold to break even for the cost of the booth rental.

Andrea sells photographs at art fairs. She prices the photos according to size: small photos cost [latex]$10[/latex], medium photos cost [latex]$15[/latex], and large photos cost [latex]$40[/latex]. She usually sells as many small photos as medium and large photos combined. She also sells twice as many medium photos as large. A booth at the art fair costs [latex]$300[/latex].

If her sales go as usual, how many of each size photo must she sell to pay for the booth?

To set up the system, first choose the variables. In this case the unknown values are the number of small, medium, and large photos.

S = number of small photos sold

M = number of medium photos sold

L = number of large photos sold

The total of her sales must be [latex]$300[/latex] to pay for the booth. We can find the total by multiplying the cost for each size by the number of that size sold.

[latex]10[/latex] S = money received for small photos

[latex]15[/latex] M = money received for medium photos

[latex]40[/latex] L = money received for large photos

Total Sales:[latex]10[/latex] S +[latex]15[/latex] M +[latex]40[/latex] L =[latex]300[/latex]

The number of small photos is the same as the total of medium and large photos combined.

She sells twice as many medium photos as large photos.

M =[latex]2[/latex]L

To make things easier, rewrite the equations to be in the same format, with all variables on the left side of the equal sign and only a constant number on the right.

[latex]\begin{cases}10S+15M+40L=300\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,S–M–L=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M–2L=0\end{cases}[/latex]

Now solve the system.

Step 1: First choose two equations and eliminate a variable. Since one equation has no S variable, it may be helpful to use the other two equations and eliminate the S variable from them. Multiply both sides of the second equation by [latex]−10[/latex].

[latex]\begin{array}{l}-10(S–M–L)=-10(0)\\-10s+10M+10L=0\end{array}[/latex]

Now add this modified equation with the first equation in the original list of equation.

[latex]\begin{array}{ccc}10S+15M+40L=300\\\underline{+(-10s+10M+10L=0)}\\25M+50L=300\end{array}[/latex]

Step 2:  The other equation for our two-variable system will be the remaining equation (that has no S variable). Eliminate a second variable using the equation from step [latex]1[/latex]. While you could multiply the third of the original equations by [latex]25[/latex] to eliminate L , the numbers will stay nicer if you divide the resulting equation from step [latex]1[/latex] by [latex]25[/latex]. Do not forget to be careful of the signs!

Divide first:

[latex]\begin{array}{ccc}\dfrac{25}{25}M+\dfrac{50}{25}L=\dfrac{300}{25}\\M+2L=12\end{array}[/latex]

Now eliminate L by adding M-2L=0 to this new equation.

[latex]\begin{array}{l}M+2L=12\\\underline{M–2L=0}\\2M=12\\M=\dfrac{12}{2}=6\end{array}[/latex]

Step 3: Use M=[latex]6[/latex] and one of the equations containing just two variables to solve for the second variable.  It is best to use one of the original equation in case an error was made in multiplication.

[latex]\begin{array}{ccc}M-2L=0\\6-2L=0\\-2L=-6\\L=3\end{array}[/latex]

Step 4: Use the two found values and one of the original equations to solve for the third variable.

[latex]\begin{array}{ccc}S–M–L=0\\S-6-3=0\\S-9=0\\S=9\end{array}[/latex]

Step 5:  Check your answer . With application problems, it is sometimes easier (and better) to use the original wording of the problem rather than the equations you write.

She usually sells as many small photos as medium and large photos combined.

  • Medium and large photos combined: [latex]6 + 3 = 9[/latex], which is the number of small photos.

She also sells twice as many medium photos as large.

  • Medium photos is [latex]6[/latex], which is twice the number of large photos [latex](3)[/latex].

A booth at the art fair costs   [latex]$300[/latex].

  • Andrea receives [latex]$10(9)[/latex] or [latex]$90[/latex] for the [latex]9[/latex] small photos, [latex]$15(6)[/latex] or [latex]$90[/latex] for the [latex]6[/latex] medium photos, and [latex]$40(3)[/latex] or [latex]$120[/latex] for the large photos. [latex]$90 + $90 + $120 = $300[/latex].

If Andrea sells [latex]9[/latex] small photos, [latex]6[/latex] medium photos, and [latex]3[/latex] large photos, she will receive exactly the amount of money needed to pay for the booth.

In the following video example, we show how to define a system of three equations in three variables that represents a mixture needed by a chemist.

Our last example shows you how to write a system of three equations that represents ticket sales for a theater that has three different prices for tickets.

  • Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
  • College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
  • Question ID 23765. Authored by : Shahbazian,Roy. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
  • Systems of Equations in Three Variables: Part 1 of 2 . Authored by : Sousa, James (Mathispower4u.com). Located at : https://youtu.be/wIE8KSpb-E8 . License : CC BY: Attribution
  • System of 3 Equations with 3 Unknowns Application - Concentration Problem. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/612Ad0W9ZeY . License : CC BY: Attribution
  • System of 3 Equations with 3 Unknowns Application - Ticket Sales. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/Wg_v5R7BFo0 . License : CC BY: Attribution

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Solving a Linear System of Linear Equations in Three Variables by Substitution

The substitution method involves algebraic substitution of one equation into a variable of the other.

This will be the sample equation used through out the instructions:

Equation 1) x – 6y – 2z = -8

Equation 2) -x + 5y + 3z = 2

Equation 3) 3x - 2y – 4z = 18

Steps in order to solve systems of linear equations through substitution:

  • Solve one of the equations for one of its variables. From the three variables, there is no incorrect choice so choose to solve for any variable.

ü       i.e.: x= 6y +2z -8

  • Next, substitute the value from the first variable you solved for into the other equation and solve for the next variable.

ü       i.e.: -(6y + 2z -8) +2y-5z = -30

-y + z + 8 = 2

  • Substitute the value from the two variables that you solved and plug it into the remaining equation and solve for the last remaining variable. This step should allow you to solve for a real number.

ü       i.e.: 3(6y + 2y – 8) – 2y – 4 (y – 6) = 18

18y – 36 = 18

  • After solving for the final variable, plug in the value of the most recent variable that you found (in terms of the example, y=3) into the answer of another equations with variables remaining (in terms of the example, z = y – 6, x = 6). Note: Preferably, plug in the value to the most simplified equation.

ü       i.e.: z = 3 – 6

  • After solving for another variable, you should have the remaining pieces of the puzzle for the last equation.

ü       i.e.: x = 6(3) + 2(-3) – 8

x = 18 – 6 – 8

  • Therefore, in the end, you will have successfully have found the answers to a system of linear equations in three variables.

ü       Answer = (4, 3, -3)

PRACTICE PROBLEMS

Practice and fine-tune your substitution skills!

  • -2x + y + 6z = 1

3x + 2y + 5z = 16

7x + 3y – 4z = 11

  • x – 3y + 6z = 21

3x + 2y -5z = -30

            2x – 5y + 2z = -6

  • x – 6y – 2z = -8

-x + 5y + 3y = 2

3x – 2y – 4z = 18

Challenge Problem

Hint: Stay open-minded

                 

      3x + 3y +z = 30

        10x - 3y - 7z = 17

-6y + 7y + 3z = -49

SOLVING LINEAR EQUATIONS IN THREE VARAIBLES PRACTICE PROBLEMS

Problem 1 :

Solve the equations

x+2y+3z  =  14, 3x+y+2z  =  11, 2x+3y+z  =  11

x+2y+3z  =  14  --------(1)

3x+y+2z  =  11 --------(2)

2x+3y+z  =  11  --------(3)

x+2y+3z - 3(2x+3y+z)  =  14 - 3(11)

x+2y+3z - 6x-9y-3z  =  14 - 33

-5x-7y  =  -19 ----(4)

3x+y+2z-2(2x+3y+z)  =  11-2(11)

3x+y+2z-4x-6y-2z  =  11-22

-x-5y  =  -11 ----(5)

-5x-7y-5(-x-5y)  =  -19-5(-11)

-5x-7y+5x+25y  =  -19+55

-7y+25y  =  36

18y  =  36

y  =  2

By applying the value of y in (5), we get

-x-5(2)  =  -11

-x-10  =  -11

-x  =  -1

x  =  1

By applying the value of x and y in (1), we get

1+2(2)+3z  =  14

5+3z =  14

3z  =  9

z  =  3

So, the solution is x  =  1, y  =  2 and z  =  3.

Problem 2 :

3x-3y+4z  =  14, -9x-6y+2z  =  1, 6x+3y+z  =  5

3x-3y+4z  =  14 ----(1)

-9x-6y+2z  =  1  ----(2)

6x+3y+z  =  5  ----(3)

2(3x-3y+4z)-( -9x-6y+2z)  =  2(14) - 1

6x-6y+8z+9x+6y-2z  =  28-1

15x+6z  =  27  -----(4)

-9x-6y+2z+ 2( 6x+3y+z )  =  1 + 2 ⋅5

-9x-6y+2z+12x+6y+2z  =  1 + 10

3x+4z  =  11   -----(5)

2(4) - 3(5)

2(15x+6z)-3(3x+4z)  =  2 ⋅ 27 - 3 ⋅ 11

30x+12z-9x-12z  =  54-33

21x  =  21

By applying the value of x in (5), we get

3(1)+4z  =  11

3+4z  =  11

4z  =  8

z  =  2

By applying the value of x and z in (1), we get

3(1)-3y+4(2)  =  14

3-3y+8  =  14

-3y+11  =  14

-3y  =  3

y  =  -1

Therefore solution is x  =  1, y  =  -1 and z  =  2

Problem 3 :

3x-2y+z  =  0, 4x+6y-3z  =  13, x-2y+2z  =  -4

3x-2y+z  =  0  -----(1)

4x+6y-3z  =  13   -----(2)

x-2y+2z  =  -4   -----(3)

3(3x-2y+z)+( 4x+6y-3z)  =  0+13

9x-6y+3z+4x+6y-3z  =  13

9x+4x  =  13

13x  =  13

(4x+6y-3z)+3(x-2y+2z)  =  13+3⋅(-4)

4x+6y-3z+3x-6y+6z  =  13-12

7x+3z  =  1 ----(4)

By applying the value of x in (4), we get

7(1)+3z  =  1

3z  =  -6

z  =  -2

3(1)-2y-2  =  0 

3-2-2y  =  0

-2y  =  -1

y  =  1/2

Therefore solution is x  =  1, y  =  1/2 and z  =  -2

Problem 4 :

2x-2y+4z  =  -12, 3x+2y+2z  =  19, -x+y-z  =  3

2x-2y+4z  =  -12 -----(1)

3x+2y+2z  =  19   -----(2)

-x+y-z  =  3    -----(3)

(2x-2y+4z)+(3x+2y+2z)  =  -12+19

5x+6z  =  7 ------(4)

(3x+2y+2z)-2( -x+y-z)  =  19-2 ⋅ 3

3x+2y+2z+2x-2y+2z  =  19-6

5x+4z  =  13   ------(5)

5x+6z - (5x+4z)  =  7-13

2z  =  -6

z  =  -3

By applying the value of z in (4), we get

5x+4(-3)  =  13

5x  =  13+12

5x  =  25

x  =  5

By applying the value of x and z in (3), we get

-5+y-(-3)  =  3

-5+y+3  =  3

-2+y  =  3

y  =  5

Therefore solution is x  =  5, y  =  5 and z  =  -3.

Problem 5 :

2x+3y-z  =  5, 4x+y+3z  =  5, 3x+2y+2z  =  5

2x+3y-z  =  5 ----(1)

4x+y+3z  =  5 ----(2)

3x+2y+2z  =  5 ----(3)

(2x+3y-z)-3( 4x+y+3z)  =  5-3 ⋅ 5

2x-12x+3y-3y-z-9z  =  5 - 15

-10x-10z  =  -10

x+z  =  1----(4)

2(4x+y+3z)-( 3x+2y+2z)  =  2 ⋅ 5-5

8x+2y+6z -3x-2y-2z  =  5

5x+4z  =  5 ---(5)

4x+4z-5x-4z  =  4-5

By applying the value of x in (1), we get

5(1)+4z  =  5

4z  =  0

z  =  0

By applying the value of z and x in (1), we get

2(1)+3y-0  =  5

3y  =  3

y  =  1

Therefore solution is x  =  1, y  =  1 and z  =  0.

Problem 6 :

x+2y+3z  =  10, x-2y+4z  =  3, x+y–3z  =  2

x+2y+3z  =  10  ----(1)

x-2y+4z  =  3   ----(2)

x+y–3z  =  2    ----(3)

(x+2y+3z)+( x-2y+4z)  =  10+3

2x+7z  =  13  ----(4)

(x-2y+4z)+2( x+y–3z)  =  3+4

x+2x-2y+2y+4z-6z  =  7

3x-2z  =  7   ----(5)

2(2x+7z)+7( 3x-2z)  =  26+7 ⋅7

4x+14z+21x-14z  =  26+49

25x  =  75

x  =  3

By applying the value of x in (3), we get

3(3)-2z  =  7

9-2z  =  7

-2z  =  7-9

-2z  =  -2

z  =  1

3+2y+3(1)  =  10

6+2y  =  10

2y  =  4

Therefore solution is x  =  3, y  =  2 and z  =  1.

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  1. solve three equations having three variable using // Casio fx-991MS # scientific calculator shorts

  2. Solving Equations with Variables on both sides Day 2

  3. 1.3 Solving Equations with Variables on Both Sides

  4. 1001 SOLVED PROBLEMS IN ENGINEERING MATHEMATICS

  5. Solving Multivariable Equations Three Equations and variable

  6. 4.2

COMMENTS

  1. Algebra

    Here is a set of practice problems to accompany the Linear Systems with Three Variables section of the Systems of Equations chapter of the notes for Paul Dawkins Algebra course at Lamar University. ... 9.4 Separation of Variables; 9.5 Solving the Heat Equation; 9.6 Heat Equation with Non-Zero Temperature Boundaries; 9.7 Laplace's Equation; 9.8 ...

  2. 4.4 Solve Systems of Equations with Three Variables

    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.

  3. 7.3: Systems of Linear Equations with Three Variables

    Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example \(\PageIndex{3}\). A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example \(\PageIndex{4}\).

  4. PDF Systems of Three Equations Elimination

    Critical thinking question: 17) Write a system of equations with the solution. Many answers. Ex: x + y + z = 3, 2 x + y + z = 5, x + 2 y − z = 4. Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com.

  5. Solving Systems of Equations Worksheets with Three Variables

    Solve using Any Method. Each printable worksheet in this unit of solving systems of equations offers eight sets of equations. High school students use their discretion to choose from the substitution method, elimination method or the Cramer's Rule to find the solution to the systems of equations involving 3 variables. Download the set.

  6. 6.5: Solve Systems of Equations with Three Variables

    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 \[ax+by+cz=d\nonumber \] Every solution to the equation is an ordered triple, \((x,y,z)\) that makes the equation true. How to solve a system of linear equations with three variables.

  7. 3.4: Solving Linear Systems with Three Variables

    Solve: {3x + 2y − z = − 7 (1) 6x − y + 3z = − 4 (2) x + 10y − 2z = 2 (3) Solution. All three equations are in standard form. If this were not the case, it would be a best practice to rewrite the equations in standard form before beginning this process. Step 1: Choose any two of the equations and eliminate a variable.

  8. IXL

    IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more. 0.

  9. Solving linear systems with 3 variables (video)

    Here are the steps. 1. Turn on your graphing calculator. (It needs to be a TI-83 or better) 2. click 2nd, matrix. 3. click to the right until you are on the setting, EDIT. 4. select 1 of the matrices. It will bring up the matrix size on the top row and the matrix at the bottom. 5. change the matrix size to 3 x 4.

  10. 5.4 Solving for Three Variables

    In solving systems of equations with three variables, use the strategies that are used to solve systems of two equations. ... These systems of equations require slightly more thought to solve than the previous problems. Example 5.4.2. Find the intersection or the solution to the following system of equations: [latex]x+2y-z=0, 3x-2y=-2,[/latex ...

  11. Linear Systems with Three Variables

    Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y. Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z. We can use this z value to find y. So the solution set is x = 1, y = 2, and z = -5/3.

  12. How to solve systems of 3 variable equations

    Practice Problems. Problem 1. Use elimination to solve the following system of three variable equations. A) 4x + 2y - 2z = 10. B) 2x + 8y + 4z = 32. C) 30x + 12y - 4z = 24. Solution. Problem 2. Use elimination to solve the following system of three variable equations.

  13. Intro to linear systems with 3 variables (video)

    9x + 15y - 108 = −48x −8y + 76. 57x + 23y = 184. Now we do a similar procedure using this and the third equation (the one that never had the z in it) 57x + 23y = 184 AND 9x-3y=25. Pick a variable to solve both equations for and then set them equal, which will give you just one variable.

  14. Solve Systems of Three Equations in Three Variables

    How To: Given a linear system of three equations, solve for three unknowns. Pick any pair of equations and solve for one variable. Pick another pair of equations and solve for the same variable. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system. Back-substitute known variables into any one of the ...

  15. Cramer's Rule With Three Variables

    Next, I will solve for the determinant of each matrix. To do this, I can manually solve the determinant of each matrix on paper using the formula provided above. It can be tedious, but it's okay since good math skills are developed by doing lots of problems. The values of the determinants are listed below. Determinants of each matrix:

  16. Systems of Three Equations: Problems

    Systems of Three Equations Math. Study Guide; Topics. Introduction and Summary; Solving by Addition and Subtraction; Problems; Solving using Matrices and Row Reduction; ... Problem : Solve the following system using the Addition/Subtraction method: x + y - 2z = 5 2x + 3y + 4z = 18 - x - 2y - 6z = - 13

  17. 3-variable linear system word problem (video)

    I having difficulty with word problem of system linear equation in three variables. Q: The perimeter of a triangle is 36 inches. Twice the length of the longest side minus the length of the shortest is 26 inches. the sum of the length of the longest side and twice the sum of both the other side length os 56 inches. find the side length.

  18. Solving Three Equations with Three Variables by Substitution

    Equation 1)x - 6y - 2z = -8. Equation 2) -x + 5y + 3z = 2. Equation 3) 3x - 2y - 4z = 18. Steps in order to solve systems of linear equations through substitution: Solve one of the equations for one of its variables. From the three variables, there is no incorrect choice so choose to solve for any variable. i.e.: x= 6y +2z -8.

  19. Algebra

    Not every linear system with three equations and three variables uses the elimination method exclusively so let's take a look at another example where the substitution method is used, at least partially. Example 2 Solve the following system of equations. 2x−4y +5z =−33 4x−y =−5 −2x+2y −3z =19 2 x − 4 y + 5 z = − 33 4 x − y ...

  20. Solving Linear Equations in Three Variables Practice Problems

    Solving Linear Equations in Three Variables Practice Problems. SOLVING LINEAR EQUATIONS IN THREE VARAIBLES PRACTICE PROBLEMS. Problem 1 : Solve the equations. x+2y+3z = 14, 3x+y+2z = 11, 2x+3y+z = 11 ... Problem 3 : Solve the equations. 3x-2y+z = 0, 4x+6y-3z = 13, x-2y+2z = -4. Solution : 3x-2y+z = 0 -----(1) 4x+6y ...

  21. Mastering Systems of Three Variable Equations: Practice 3-6 Answers

    In practice 3 6, students are tasked with solving systems of three variable equations and finding the answers. This requires understanding the relationships between the variables and using techniques such as substitution or elimination to find the solutions. One approach to solving these systems is to use substitution.