problem solving system of equations

4.4 Solve Systems of Equations with Three Variables

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

Section 4.4 Exercises

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?

Number Line

• x+y+z=25,\:5x+3y+2z=0,\:y-z=6
• x+2y=2x-5,\:x-y=3
• 5x+3y=7,\:3x-5y=-23
• x^2+y=5,\:x^2+y^2=7
• xy+x-4y=11,\:xy-x-4y=4
• 3-x^2=y,\:x+1=y
• xy=10,\:2x+y=1
• How do you solve a system of equations by substitution?
• To solve a system of equations by substitution, solve one of the equations for one of the variables, and substitute this expression into the other equation. Then, solve the resulting equation for the remaining variable and substitute this value back into the original equation to find the value of the other variable.
• How do you solve a system of equations by graphing?
• To solve a system of equations by graphing, graph both equations on the same set of axes and find the points at which the graphs intersect. Those points are the solutions.
• How do you solve a system of equations by elimination?
• To solve a system of equations by elimination, write the system of equations in standard form: ax + by = c, and multiply one or both of the equations by a constant so that the coefficients of one of the variables are opposite. Then, add or subtract the two equations to eliminate one of the variables. Solve the resulting equation for the remaining variable.
• What are the solving methods of a system of equations?
• There are several methods for solving a system of equations, including substitution, elimination, and graphing.
• What is a system of linear equations?
• A system of linear equations is a system of equations in which all the equations are linear and in the form ax + by = c, where a, b, and c are constants and x and y are variables.

system-of-equations-calculator

• High School Math Solutions – Systems of Equations Calculator, Nonlinear In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how...

Please add a message.

Message received. Thanks for the feedback.

Online Systems of Equations Solver

Solve equations and systems of equations with wolfram|alpha, a powerful tool for finding solutions to systems of equations and constraints.

Wolfram|Alpha is capable of solving a wide variety of systems of equations. It can solve systems of linear equations or systems involving nonlinear equations, and it can search specifically for integer solutions or solutions over another domain. Additionally, it can solve systems involving inequalities and more general constraints.

Learn more about:

• Systems of equations

Tips for entering queries

Enter your queries using plain English. To avoid ambiguous queries, make sure to use parentheses where necessary. Here are some examples illustrating how to ask about solving systems of equations.

• solve y = 2x, y = x + 10
• solve system of equations {y = 2x, y = x + 10, 2x = 5y}
• y = x^2 - 2, y = 2 - x^2
• solve 4x - 3y + z = -10, 2x + y + 3z = 0, -x + 2y - 5z = 17
• solve system {x + 2y - z = 4, 2x + y + z = -2, z + 2y + z = 2}
• solve 4 = x^2 + y^2, 4 = (x - 2)^2 + (y - 2)^2
• x^2 + y^2 = 4, y = x
• View more examples

Access instant learning tools

Get immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator

• Step-by-step solutions
• Wolfram Problem Generator

What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

Solving systems of equations is a very general and important idea, and one that is fundamental in many areas of mathematics, engineering and science.

System of Equations

In mathematics, a system of equations, also known as a set of simultaneous equations or an equation system, is a finite set of equations for which we sought common solutions. In systems of equations, variables are related in a specific way in each equation. i.e., the equations can be solved simultaneously to find a set of values of variables that satisfies each equation.

System of linear equations finds applications in our day-to-day lives in modelling problems where the unknown values can be represented in form of variables. Solving system of equations involves different methods such as substitution, elimination, graphing, etc. Let us look into each method in detail.

What is a System of Equations?

In algebra, a system of equations comprises two or more equations and seeks common solutions to the equations. "A system of linear equations is a set of equations which are satisfied by the same set of values of variables."

System of Equations Example

A system of equations as discussed above is a set of equations that seek a common solution for the variables included. The following set of linear equations is an example of the system of equations:

• 2x - y = 12
• x - 2y = 48

Note that the values x = -8 and y = -28 satisfy each of these equations and hence the pair (x, y) = (-8, -28) is the solution of the above system of equations. But how to solve system of equations? Let us see.

Solutions of System of Equations

Solution of system of equations is the set of values of variables that satisfies each linear equation in the system. The main reason behind solving an equation system is to find the value of the variable that satisfies the condition of all the given equations true. There systems of equations are classified into 3 types depending on their number of solutions:

• Linear System with "Unique solution"
• Linear System with "No solution"
• Linear System with "Infinitely many solutions"

We know that every linear equation represents a line on the coordinate plane. In this perception, the above figure should give more sense to understanding the different types of solutions of system of equations.

Unique Solution of a System of Equations

A system of equations has unique solution when there is only set of variables exist that satisfy each equation in the system. In terms of graphs, a system with a unique solution has lines (representing the equations) that intersect (at one point).

No Solution

A system of equations has no solution when there exists no set of variables that satisfy each linear equation in the system. If we graph that kind of system, the resulting lines will be parallel to each other.

Infinite Many Solutions

A system of equations can have infinitely many solutions when there exist infinite sets of variables that satisfy each equation. In such cases, the lines corresponding to the linear equations would overlap each other on the graph. i.e., both equations represent the same line. Since a line has infinite points on it, each point on it becomes a solution of the system.

Solving System of Equations

Solving a system of equations means finding the values of the variables used in the set of equations. Any system of equations can be solved in different methods.

• Substitution Method
• Elimination Method
• Graphical Method
• Cross-multiplication method

To solve a system of equations in 2 variables, we need at least 2 equations. Similarly, for solving a system of equations in 3 variables, we will require at least 3 equations. Let us understand 3 ways to solve a system of equations given the equations are linear equations in two variables.

☛ Also Check: Solving System of Linear Equations

Solving System of Linear Equations By Substitution Method

For solving the system of equations using the substitution method given two linear equations in x and y, in one of the equations, express y in terms x in one of the equations and then substitute it in the other equation.

Example: Solve the system of equations using the substitution method.

3x − y = 23 → (1)

4x + 3y = 48 → (2)

From (1), we get:

y = 3x − 23 → 3

Plug in y in (2),

4x + 3 (3x − 23) = 48

13x − 69 = 48

Now, plug in x = 9 in (1)

y = 3 × 9 − 23 = 4

Hence, x = 9 and y = 4 is the solution of the given system of equations.

Solving System of Equations By Elimination Method

Using the elimination method to solve the system of equations, we eliminate one of the unknowns, by multiplying equations by suitable numbers, so that the coefficients of one of the variables become the same.

Example: Solve the following system of linear equations by elimination method.

2x + 3y = 4 → (1) and 3x + 2y = 11 → (2)

The coefficients of y are 3 and 2; LCM (3, 2) = 6

Multiplying Equation (1) by 2 and Equation (2) by 3, we get

4x + 6y = 8 → (3)

9x + 6y = 33 → (4)

On subtracting (3) from (4), we get

Plugging in x = 5 in (2) we get

15 + 2y = 11

Hence, x = 5, y = −2 is the solution.

Solving System of Equations by Graphing

In this method, solving system of linear equations is done by plotting their graphs. "The point of intersection of the two lines is the solution of the system of equations using graphical method ."

Example: 3x + 4y = 11 and -x + 2y = 3

Find at least two values of x and y satisfying equation 3x + 4y = 11

So we have 2 points A (1, 2) and B (3, 0.5).

Similarly, find at least two values of x and y satisfying the equation -x + 2y = 3

We have two points C(-3, 0) and D( 3, 3). Plotting these points on the graph we can get the lines in a coordinate plane as shown below.

We observe that the two lines intersect at (1, 2). So, x = 1, y = 2 is the solution of given system of equations. Methods I and II are the algebraic way of solving simultaneous equations and III is the graphical method .

Solving System of Equations Using Cross-multiplication method

In this method, we solve the systems of equations a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 using the cross-multiplication formula:

x / (b 1 c 2 - b 2 c 1 ) = -y / (a 1 c 2 - a 2 c 1 ) = 1 / (a 1 b 2 - a 2 b 1 )

For more detailed information of this method, click here .

Solving System of Equations Using Matrices

The solution of a system of equations can be solved using matrices . In order to solve a system of equations using matrices, express the given equations in standard form , with the variables and constants on respective sides. for the given equations,

a\(_1\)x + \(b_1\)y + \(c_1\)z = \(d_1\)

a\(_2\)x + \(b_2\)y + \(c_2\)z = \(d_2\)

a\(_3\)x + \(b_3\)y + \(c_3\)z = \(d_3\)

we can express them in the form of matrices as,

\(\left[\begin{array}{ccc} a_1x + b_1y + c_1 z \\ a_2x + b_2y + c_2 z \\ a_3x + b_3y + c_3 z \end{array}\right] = \left[\begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right]\)

⇒\(\left[\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right] + \left[\begin{array}{ccc} x \\ y \\ z \end{array}\right] = \left[\begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right]\)

A = \(\left[\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right]\), X = \(\left[\begin{array}{ccc} x \\ y \\ z \end{array}\right]\), B = \(\left[\begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right]\)

⇒ The solution of the system is given by the formula X = A -1 B, where A -1 = inverse of the matrix A. To understand this method more in detail, click here .

Alternatively, Cramer's rule can also be used to solve the system of equations using determinants .

Applications of System of Equations

Systems of equations are a very useful tool and find application in our day-to-day lives for modeling real-life situations and analyzing questions about them.

• System of Equations Word Problems
• Systems of Equations Worksheet

For applying the concept of the system of equations, we need to translate the given situation into two linear equations in two variables , then further solve to find the solution of linear programming problems. Any method to solve the system of equations, substitution, elimination, graphical, etc methods. Follow the below-given steps to apply the system of equations to solve problems in our daily lives,

• To translate and represent the given situation in form of a system of equations, identify unknown quantities in a problem and represent them with variables.
• Write a system of equations modelling the conditions of the problem.
• Solve the system of equations.
• Check and express the obtained solution in terms of the given context.

☛Related Articles:

Check these articles related to the concept of the system of equations.

• Solutions of a Linear Equation
• Simultaneous Linear Equations
• Solving Linear Equations Calculator
• Equation Calculator
• System of Equations Calculator

System of Equations Examples

Example 1: In a ΔLMN, ∠N = 3∠M = 2(∠L+∠M). Calculate the angles of ΔLMN using this information.

Let ∠L = x° and ∠M = y°.

∠N = 3∠M = (3y)°.

∠L + ∠M + ∠N = 180° ∴ x + y + 3y = 180 x + 4y = 180 ... (1) From the given equation, 3∠M = 2(∠L+∠M) ∴ 3y = 2(x + y) ⇒ 2x - y = 0 ... (2) On multiplying (2) by 4 and adding the result to (1), we get 9x = 180 x = 20 Substitute this in (2): 40 - y = 0 ⇒ y = 40 Therefore, ∠L = 20° and ∠M = 40°. Therefore, the third angle is, ∠N = 180 - (20 + 40) = 120°. Answer: ∠L = 20°, ∠M = 40°, and ∠N = 120°.

Example 2: Peter is three times as old as his son. 5 years later, he shall be two and half times as old as his son. What's Peter's present age?

Let Peter's age be x years and his son's age be y years. Then by given info,

x = 3y ... (1)

5 years later,

Peter's age = x + 5 years and his son's age = y + 5 years.

By the given condition,

x + 5 = 5/2 (y + 5)

2x - 5y - 15 = 0 ... (2)

From (1), we have x = 3y. Substitute this in (2):

2(3y) - 5y - 15 = 0

Substitute this in (1): x = 3(15) = 45

Thus, the solution of the given system is (x, y) = (45, 15).

Answer: Present age of Peter = 45 years; Present age of son = 15 years.

Example 3: Tressa starts her job with a certain monthly salary and earns a fixed increment every year. If her salary was $1500 after 4 years and $1800 after 10 years of service, find her starting salary and annual increment.

Let her starting salary be x and annual increment be y.

After 4 years, her salary was $1500

x + 4y = 1500 ... (1)

After 10 years, her salary was $1800

x + 10y = 1800 ... (2)

Subtracting Eqn (1) from (2), we get

On putting y = 50 in (1), we get x = 1300.

Answer: Starting salary was $1300 and annual increment is $50.

go to slide go to slide go to slide

Book a Free Trial Class

Practice Questions on System of Equations

go to slide go to slide

FAQs on System of Equations

What is a system of equations in mathematics.

A system of equations in mathematics is a set of linear equations that need to be solved to find a common solution. A real-life problem with two or more unknowns can be converted into a system of equations and can be solved to find a set of values of variables that satisfy all equations.

How Do You Solve a System of Equations?

Solving a system of equations is computing the unknown variables still balancing the equations on both sides. We solve an equation system to find the values of the variables that satisfy the condition of all the given equations. There are different methods to solve a system of equations,

• Graphical method
• Substitution method
• Elimination method

How Do You Create a System of Equations with Two Variables?

To create a system of equations with two variables:

• First, identify the two unknown quantities in the given problem.
• Next, find out the two conditions given and frame equations for each of them.

How to Solve a System of Equations by Substitution Method?

The substitution method is one of the ways to solve a system of equations in two variables, given the set of linear equations. In this method, we substitute the value of a variable found by one equation in the second equation.

How to Solve a System of Equations Using the Elimination Method?

The elimination method is used to solve a system of linear equations. In the elimination method, we eliminate one of the two variables by multiplying each equation with the required numbers and try to solve equations with another variable. In this process, finding LCM of coefficients would be helpful.

What is the Purpose of Graphing Systems of Equations?

To solve the system of equations, given a set of linear equations graphically, we need to find at least two solutions for the respective equations. We observe the pattern of lines after plotting the points to infer it is consistent, dependent, or inconsistent.

• If the two lines are intersecting at the same point, then the intersection point gives a unique solution for the system of equations.
• If the two lines coincide, then in this case there are infinitely many solutions.
• If the two lines are parallel, then in this case there is no solution.

What are Homogeneous System of Linear Equations?

The homogeneous system of linear equations is a set of linear equation each one of which has its constant term to be 0. The process of solving these kind of systems can be learnt in detailed by clicking here .

How to Solve a System of Equations With Cross Multiplication Method?

While solving a system of equations using the cross-multiplication method , we use the formula x / (b 1 c 2 - b 2 c 1 ) = -y / (a 1 c 2 - a 2 c 1 ) = 1 / (a 1 b 2 - a 2 b 1 ) to solve the system a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0.

How do you Solve a System of Equations Using 2 Equations with 3 Variables?

An equation with 3 variables represents a plane .

• Step 1) To solve a system of 2 equations with 3 variables say x, y, and z, we will consider the 1st two equations and eliminate one of the variables, say x, to obtain a new equation.
• Step 2) Next, we write the 2nd variable, y in terms of z from the new equation and substitute it in the third equation.
• Step 3) Assuming z = a, we will obtain values of x and y also in terms of 'a'.
• Step 4) Once, we know the value of a, we can find the values of x and y in terms of 'a'.

What is the Solution to the System of Equations and How to Find It?

Solving system of equations is finding the values of variables that satisfy each equation in that system. There are three main methods to solving system of equations, they are:

Where Can I Find System of Linear Equations Calculator?

The system of linear equations calculator is available here . This allows us to enter the linear equations. Then it will show the solution along with step-by-step solution.

Systems of Linear Equations

A linear equation is not always in the form y = 3.5 − 0.5x ,

It can also be like y = 0.5(7 − x)

Or like y + 0.5x = 3.5

Or like y + 0.5x − 3.5 = 0 and more.

(Note: those are all the same linear equation!)

A System of Linear Equations is when we have two or more linear equations working together.

Example: Here are two linear equations:

Together they are a system of linear equations.

Can you discover the values of x and y yourself? (Just have a go, play with them a bit.)

Let's try to build and solve a real world example:

Example: You versus Horse

It's a race!

You can run 0.2 km every minute.

The Horse can run 0.5 km every minute. But it takes 6 minutes to saddle the horse.

How far can you get before the horse catches you?

We can make two equations ( d =distance in km, t =time in minutes)

• You run at 0.2km every minute, so d = 0.2t
• The horse runs at 0.5 km per minute, but we take 6 off its time: d = 0.5(t−6)

So we have a system of equations (that are linear ):

• d = 0.5(t−6)

We can solve it on a graph:

Do you see how the horse starts at 6 minutes, but then runs faster?

It seems you get caught after 10 minutes ... you only got 2 km away.

Run faster next time.

So now you know what a System of Linear Equations is.

Let us continue to find out more about them ....

There can be many ways to solve linear equations!

Let us see another example:

Example: Solve these two equations:

• −3x + y = 2

The two equations are shown on this graph:

Our task is to find where the two lines cross.

Well, we can see where they cross, so it is already solved graphically.

But now let's solve it using Algebra!

Hmmm ... how to solve this? There can be many ways! In this case both equations have "y" so let's try subtracting the whole second equation from the first:

Now let us simplify it:

So now we know the lines cross at x=1 .

And we can find the matching value of y using either of the two original equations (because we know they have the same value at x=1). Let's use the first one (you can try the second one yourself):

And the solution is:

x = 1 and y = 5

And the graph shows us we are right!

Linear Equations

Only simple variables are allowed in linear equations. No x 2 , y 3 , √x, etc :

Common Variables

Equations that "work together" share one or more variables:

A System of Equations has two or more equations in one or more variables

Many Variables

So a System of Equations could have many equations and many variables.

Example: 3 equations in 3 variables

There can be any combination:

• 2 equations in 3 variables,
• 6 equations in 4 variables,
• 9,000 equations in 567 variables,

When the number of equations is the same as the number of variables there is likely to be a solution. Not guaranteed, but likely.

In fact there are only three possible cases:

• No solution
• One solution
• Infinitely many solutions

When there is no solution the equations are called "inconsistent" .

One or infinitely many solutions are called "consistent"

Here is a diagram for 2 equations in 2 variables :

Independent

"Independent" means that each equation gives new information. Otherwise they are "Dependent" .

Also called "Linear Independence" and "Linear Dependence"

• 2x + 2y = 6

Those equations are "Dependent" , because they are really the same equation , just multiplied by 2.

So the second equation gave no new information .

Where the Equations are True

The trick is to find where all equations are true at the same time .

True? What does that mean?

The "you" line is true all along its length (but nowhere else).

Anywhere on that line d is equal to 0.2t

• at t=5 and d=1, the equation is true (Is d = 0.2t? Yes, as 1 = 0.2×5 is true)
• at t=5 and d=3, the equation is not true (Is d = 0.2t? No, as 3 = 0.2×5 is not true )

Likewise the "horse" line is also true all along its length (but nowhere else).

But only at the point where they cross (at t=10, d=2) are they both true .

So they have to be true simultaneously ...

... that is why some people call them "Simultaneous Linear Equations"

Solve Using Algebra

It is common to use Algebra to solve them.

Here is the "Horse" example solved using Algebra:

The system of equations is:

In this case it seems easiest to set them equal to each other:

d = 0.2t = 0.5(t−6)

Now we know when you get caught!

And our solution is:

t = 10 minutes and d = 2 km

Algebra vs Graphs

Why use Algebra when graphs are so easy? Because:

More than 2 variables can't be solved by a simple graph.

So Algebra comes to the rescue with two popular methods:

Solving By Substitution

Solving by elimination.

We will see each one, with examples in 2 variables, and in 3 variables. Here goes ...

These are the steps:

• Write one of the equations so it is in the style "variable = ..."
• Replace (i.e. substitute) that variable in the other equation(s).
• Solve the other equation(s)
• (Repeat as necessary)

Here is an example with 2 equations in 2 variables :

• 3x + 2y = 19

We can start with any equation and any variable .

Let's use the second equation and the variable "y" (it looks the simplest equation).

Write one of the equations so it is in the style "variable = ...":

We can subtract x from both sides of x + y = 8 to get y = 8 − x . Now our equations look like this:

Now replace "y" with "8 − x" in the other equation:

• 3x + 2 (8 − x) = 19

Solve using the usual algebra methods:

Expand 2(8−x) :

• 3x + 16 − 2x = 19

Then 3x−2x = x :

• x + 16 = 19

And lastly 19−16=3

Now we know what x is, we can put it in the y = 8 − x equation:

• y = 8 − 3 = 5

And the answer is:

x = 3 y = 5

Note: because there is a solution the equations are "consistent"

Check: why don't you check to see if x = 3 and y = 5 works in both equations?

Solving By Substitution: 3 equations in 3 variables

OK! Let's move to a longer example: 3 equations in 3 variables .

This is not hard to do... it just takes a long time !

• 2x + y + 3z = 15

We should line up the variables neatly, or we may lose track of what we are doing:

We can start with any equation and any variable. Let's use the first equation and the variable "x".

Now replace "x" with "6 − z" in the other equations:

(Luckily there is only one other equation with x in it)

2(6−z) + y + 3z = 15 simplifies to y + z = 3 :

Good. We have made some progress, but not there yet.

Now repeat the process , but just for the last 2 equations.

Let's choose the last equation and the variable z:

Now replace "z" with "3 − y" in the other equation:

−3y + (3−y) = 7 simplifies to −4y = 4 , or in other words y = −1

Almost Done!

Knowing that y = −1 we can calculate that z = 3−y = 4 :

And knowing that z = 4 we can calculate that x = 6−z = 2 :

x = 2 y = −1 z = 4

Check: please check this yourself.

We can use this method for 4 or more equations and variables... just do the same steps again and again until it is solved.

Conclusion: Substitution works nicely, but does take a long time to do.

Elimination can be faster ... but needs to be kept neat.

"Eliminate" means to remove : this method works by removing variables until there is just one left.

The idea is that we can safely :

• multiply an equation by a constant (except zero),
• add (or subtract) an equation on to another equation

Like in these two examples:

CAN we safely add equations to each other?

Yes, because we are "keeping the balance".

Imagine two really simple equations:

x − 5 = 3 5 = 5

We can add the "5 = 5" to "x − 5 = 3":

x − 5 + 5 = 3 + 5 x = 8

Try that yourself but use 5 = 3+2 as the 2nd equation

It works just fine, because both sides are equal (that is what the = is for)

We can also swap equations around, so the 1st could become the 2nd, etc, if that helps.

OK, time for a full example. Let's use the 2 equations in 2 variables example from before:

Very important to keep things neat:

Now ... our aim is to eliminate a variable from an equation.

First we see there is a "2y" and a "y", so let's work on that.

Multiply the second equation by 2:

Subtract the second equation from the first equation:

Yay! Now we know what x is!

Next we see the 2nd equation has "2x", so let's halve it, and then subtract "x":

Multiply the second equation by ½ (i.e. divide by 2):

Subtract the first equation from the second equation:

x = 3 and y = 5

And here is the graph:

The blue line is where 3x + 2y = 19 is true

The red line is where x + y = 8 is true

At x=3, y=5 (where the lines cross) they are both true. That is the answer.

Here is another example:

• 6x − 3y = 3

Lay it out neatly:

Multiply the first equation by 3:

0 − 0 = 9 ???

What is going on here?

Quite simply, there is no solution.

And lastly:

• 6x − 3y = 12

Well, that is actually TRUE! Zero does equal zero ...

... that is because they are really the same equation ...

... so there are an Infinite Number of Solutions

And so now we have seen an example of each of the three possible cases:

Solving By Elimination: 3 equations in 3 variables

Before we start on the next example, let's look at an improved way to do things.

Follow this method and we are less likely to make a mistake.

First of all, eliminate the variables in order :

• Eliminate x s first (from equation 2 and 3, in order)
• then eliminate y (from equation 3)

Start with:

Eliminate in this order:

We then have this "triangle shape":

Now start at the bottom and work back up (called "Back-Substitution") (put in z to find y , then z and y to find x ):

And we are solved:

ALSO, it is easier to do some of the calculations in our head, or on scratch paper, instead of always working within the set of equations:

• x + y + z = 6
• 2y + 5z = −4
• 2x + 5y − z = 27

Written neatly:

First, eliminate x from 2nd and 3rd equation.

There is no x in the 2nd equation ... move on to the 3rd equation:

Subtract 2 times the 1st equation from the 3rd equation (just do this in your head or on scratch paper):

And we get:

Next, eliminate y from 3rd equation.

We could subtract 1½ times the 2nd equation from the 3rd equation (because 1½ times 2 is 3) ...

... but we can avoid fractions if we:

• multiply the 3rd equation by 2 and
• multiply the 2nd equation by 3

and then do the subtraction ... like this:

And we end up with:

We now have that "triangle shape"!

Now go back up again "back-substituting":

We know z , so 2y+5z=−4 becomes 2y−10=−4 , then 2y=6 , so y=3 :

Then x+y+z=6 becomes x+3−2=6 , so x=6−3+2=5

x = 5 y = 3 z = −2

Please check this for yourself, it is good practice.

General Advice

Once you get used to the Elimination Method it becomes easier than Substitution, because you just follow the steps and the answers appear.

But sometimes Substitution can give a quicker result.

• Substitution is often easier for small cases (like 2 equations, or sometimes 3 equations)
• Elimination is easier for larger cases

And it always pays to look over the equations first, to see if there is an easy shortcut ... so experience helps ... so get experience!

• Pre-Algebra Topics
• Algebra Topics
• Algebra Calculator
• Algebra Cheat Sheet
• Algebra Practice Test
• Algebra Readiness Test
• Algebra Formulas
• Want to Build Your Own Website?

Sign In / Register

Solving Systems of Equations Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations . Then we moved onto solving systems using the Substitution Method . In our last lesson we used the Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world problems! Are you ready? First let's look at some guidelines for solving real world problems and then we'll look at a few examples.

Steps For Solving Real World Problems

• Highlight the important information in the problem that will help write two equations.
• Define your variables
• Write two equations
• Use one of the methods for solving systems of equations to solve.
• Check your answers by substituting your ordered pair into the original equations.
• Answer the questions in the real world problems. Always write your answer in complete sentences!

Ok... let's look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

1.  Let's start by identifying the important information:

• hot dogs cost $1.50
• Sodas cost $0.50
• Made a total of $78.50
• Sold 87 hot dogs and sodas combined

2.  Define your variables.

• Ask yourself, "What am I trying to solve for? What don't I know?

In this problem, I don't know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50    (Equation related to cost)

 x + y = 87   (Equation related to the number sold)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6.  Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7.  Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn't too bad, was it? The hardest part is writing the equations. From there you already know the strategies for solving. Think carefully about what's happening in the problem when trying to write the two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend's bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

• 3 soft tacos + 3 burritos cost $11.25
• 4 soft tacos + 2 burritos cost $10.00

In this problem, I don't know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

One equation will be related your lunch and one equation will be related to your friend's lunch.

3x + 3y = 11.25  (Equation representing your lunch)

4x + 2y = 10   (Equation representing your friend's lunch)

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the whole reason why we are learning these skills. You must be able to apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

Problem about the WNBA

Systems problem about ages

Problem about milk consumption in the U.S.

Vans and Buses? How many rode in each?

Telephone Plans problem

Systems problem about hats and scarves

Apples and guavas please!

How much did Alice spend on shoes?

All about stamps

Going to the movies

Small pitchers and large pitchers - how much will they hold?

Chickens and dogs in the farm yard

• System of Equations
• Systems Word Problems

Need More Help With Your Algebra Studies?

Get access to hundreds of video examples and practice problems with your subscription! 

Click here for more information on our affordable subscription options.

Not ready to subscribe?  Register for our FREE Pre-Algebra Refresher course.

ALGEBRA CLASS E-COURSE MEMBERS

Click here for more information on our Algebra Class e-courses.

Need Help? Try This Online Calculator!

Affiliate Products...

On this site, I recommend only one product that I use and love and that is Mathway   If you make a purchase on this site, I may receive a small commission at no cost to you. 

Privacy Policy

Let Us Know How we are doing!

  send us a message to give us more detail!

Would you prefer to share this page with others by linking to it?

• Click on the HTML link code below.
• Copy and paste it, adding a note of your own, into your blog, a Web page, forums, a blog comment, your Facebook account, or anywhere that someone would find this page valuable.

Copyright © 2009-2020   |   Karin Hutchinson   |   ALL RIGHTS RESERVED.

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• This Or That Game
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics

How to Solve Systems of Equations

Last Updated: March 26, 2024 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 241,159 times.

Solving a system of equations requires you to find the value of more than one variable in more than one equation. You can solve a system of equations through addition, subtraction, multiplication, or substitution. If you want to know how to solve a system of equations, just follow these steps.

Solve by Subtraction

• For example, if both equations have the variable positive 2x, you should use the subtraction method to find the value of both variables.
• Write one equation above the other by matching up the x and y variables and the whole numbers. Write the subtraction sign outside the quantity of the second system of equations.
• 2x + 4y = 8
• -(2x + 2y = 2)

• 2x - 2x = 0
• 4y - 2y = 2y
• 2x + 4y = 8 -(2x + 2y = 2) = 0 + 2y = 6

• Divide 2y and 6 by 2 to get y = 3

• Plug y = 3 into the equation 2x + 2y = 2 and solve for x.
• 2x + 2(3) = 2
• You have solved the system of equations by subtraction. (x, y) = (-2, 3)

• 2(-2) + 4(3) = 8
• -4 + 12 = 8
• 2(-2) + 2(3) = 2

Solve by Addition

• Write one equation above the other by matching up the x and y variables and the whole numbers. Write the addition sign outside the quantity of the second system of equations.
• 3x + 6y = 8
• +(x - 6y = 4)

• 3x + x = 4x
• 6y + -6y = 0
• = 4x + 0 = 12

• 4x + 0 = 12
• Divide 4x and 12 by 3 to get x = 3

• Plug x = 3 into the equation x - 6y = 4 to solve for y.
• You have solved the system of equations by addition. (x, y) = (3, -1/6)

• 3(3) + 6(-1/6) = 8
• 3 - (6 * -1/6) =4
• 3 - - 1 = 4

Solve by Multiplication

• 3x + 2y = 10

• 2 (2x - y = 2)
• 4x - 2y = 4

• + 4x - 2y = 4
• 7x + 0 = 14

• x = 2 ---> 2x - y = 2
• You have solved the system of equations by multiplication. (x, y) = (2, 2)

• Plug (2, 2) in for (x, y) in the equation 3x + 2y = 10.
• 3(2) + 2(2) = 10
• Plug (2, 2) in for (x, y) in the equation 2x - y = 2.
• 2(2) - 2 = 2

Solve by Substitution

• If you're working with the equations 2x + 3y = 9 and x + 4y = 2, you should isolate x in the second equation.

• x = 2 - 4y --> 2x + 3y = 9
• 2(2 - 4y) + 3y = 9
• 4 - 8y + 3y = 9
• -5y = 9 - 4

• y = -1 --> x = 2 - 4y
• x = 2 - 4(-1)
• You have solved the system of equations by substitution. (x, y) = (6, -1)

• 2(6) + 3(-1) = 9
• Plug (6, -1) in for (x, y) in the equation x + 4y = 2.
• 6 + 4(-1) = 2

Community Q&A

• You should be able to solve any linear system of equations using the addition, subtraction, multiplication, or substitution method, but one method is usually the easiest depending on the equations. Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ https://content.nroc.org/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L2_T2_text_final.html
• ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
• ↑ https://www.cliffsnotes.com/study-guides/algebra/algebra-i/equations-with-two-variables/solving-systems-of-equations-simultaneous-equations
• ↑ https://www.purplemath.com/modules/systlin5.htm
• ↑ https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-8-for-ccss/section/5.5/related/lesson/solving-systems-of-equations-by-multiplication-and-addition-c-alg/
• ↑ https://virtualnerd.com/sat-math/algebra/systems-equations/equations-solution-by-elimination-multiplication
• ↑ https://www.khanacademy.org/math/algebra/systems-of-linear-equations/solving-systems-of-equations-with-substitution/v/solving-systems-with-substitution
• ↑ https://opentextbc.ca/businesstechnicalmath/chapter/solve-systems-of-equations-by-substitution/

About This Article

To solve a system of equations by elimination, make sure both equations have one variable with the same coefficient. Subtract the like terms of the equations so that you’re eliminating that variable, then solve for the remaining one. Plug the solution back into one of the original equations to solve for the other variable. To solve by substitution, solve for 1 variable in the first equation, then plug the value into the second equation and solve for the second variable. Finally, solve for the first variable in either of the first equations. Write your answer by placing both terms in parentheses with a comma between. If you want to learn how to check your answers, keep reading the article! Did this summary help you? Yes No

• Send fan mail to authors

Reader Success Stories

Did this article help you?

Dantik S. K.

May 5, 2016

Emmanuel Amas Nyonyi

Sep 23, 2017

yoshiyow Gaming and Tech

May 21, 2017

Michael Sahyoun

Nov 14, 2018

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

Don’t miss out! Sign up for

wikiHow’s newsletter

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

4.6: Solve Systems of Equations Using Matrices

• Last updated
• Save as PDF
• Page ID 5142

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

Learning Objectives

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

• Write the augmented matrix for a system of equations
• Use row operations on a matrix
• Solve systems of equations using matrices

Before you get started, take this readiness quiz.

• Solve: \(3(x+2)+4=4(2x−1)+9\). If you missed this problem, review [link] .
• Solve: \(0.25p+0.25(x+4)=5.20\). If you missed this problem, review [link] .
• Evaluate when \(x=−2\) and \(y=3:2x^2−xy+3y^2\). If you missed this problem, review [link] .

Write the Augmented Matrix for a System of Equations

Solving a system of equations can be a tedious operation where a simple mistake can wreak havoc on finding the solution. An alternative method which uses the basic procedures of elimination but with notation that is simpler is available. The method involves using a matrix . A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix is a rectangular array of numbers arranged in rows and columns.

A matrix with m rows and n columns has order \(m\times n\). The matrix on the left below has 2 rows and 3 columns and so it has order \(2\times 3\). We say it is a 2 by 3 matrix.

Each number in the matrix is called an element or entry in the matrix.

We will use a matrix to represent a system of linear equations. We write each equation in standard form and the coefficients of the variables and the constant of each equation becomes a row in the matrix. Each column then would be the coefficients of one of the variables in the system or the constants. A vertical line replaces the equal signs. We call the resulting matrix the augmented matrix for the system of equations.

Notice the first column is made up of all the coefficients of x , the second column is the all the coefficients of y , and the third column is all the constants.

Example \(\PageIndex{1}\)

ⓐ \(\left\{ \begin{array} {l} 5x−3y=−1 \\ y=2x−2 \end{array} \right. \) ⓑ \( \left\{ \begin{array} {l} 6x−5y+2z=3 \\ 2x+y−4z=5 \\ 3x−3y+z=−1 \end{array} \right. \)

ⓐ The second equation is not in standard form. We rewrite the second equation in standard form.

\[\begin{aligned} y=2x−2 \\ −2x+y=−2 \end{aligned} \nonumber\]

We replace the second equation with its standard form. In the augmented matrix, the first equation gives us the first row and the second equation gives us the second row. The vertical line replaces the equal signs.

ⓑ All three equations are in standard form. In the augmented matrix the first equation gives us the first row, the second equation gives us the second row, and the third equation gives us the third row. The vertical line replaces the equal signs.

Example \(\PageIndex{2}\)

Write each system of linear equations as an augmented matrix:

ⓐ \(\left\{ \begin{array} {l} 3x+8y=−3 \\ 2x=−5y−3 \end{array} \right. \) ⓑ \(\left\{ \begin{array} {l} 2x−5y+3z=8 \\ 3x−y+4z=7 \\ x+3y+2z=−3 \end{array} \right. \)

ⓐ \(\left[ \begin{matrix} 3 &8 &-3 \\ 2 &5 &−3 \end{matrix} \right] \)

ⓑ \(\left[ \begin{matrix} 2 &3 &1 &−5 \\ −1 &3 &3 &4 \\ 2 &8 &7 &−3 \end{matrix} \right] \)

Example \(\PageIndex{3}\)

ⓐ \(\left\{ \begin{array} {l} 11x=−9y−5 \\ 7x+5y=−1 \end{array} \right. \) ⓑ \(\left\{ \begin{array} {l} 5x−3y+2z=−5 \\ 2x−y−z=4 \\ 3x−2y+2z=−7 \end{array} \right. \)

ⓐ \(\left[ \begin{matrix} 11 &9 &−5 \\ 7 &5 &−1 \end{matrix} \right] \) ⓑ \(\left[ \begin{matrix} 5 &−3 &2 &−5 \\ 2 &−1 &−1 &4 \\ 3 &−2 &2 &−7 \end{matrix} \right] \)

It is important as we solve systems of equations using matrices to be able to go back and forth between the system and the matrix. The next example asks us to take the information in the matrix and write the system of equations.

Example \(\PageIndex{4}\)

Write the system of equations that corresponds to the augmented matrix:

\(\left[ \begin{array} {ccc|c} 4 &−3 &3 &−1 \\ 1 &2 &−1 &2 \\ −2 &−1 &3 &−4 \end{array} \right] \).

We remember that each row corresponds to an equation and that each entry is a coefficient of a variable or the constant. The vertical line replaces the equal sign. Since this matrix is a \(4\times 3\), we know it will translate into a system of three equations with three variables.

Example \(\PageIndex{5}\)

Write the system of equations that corresponds to the augmented matrix: \(\left[ \begin{matrix} 1 &−1 &2 &3 \\ 2 &1 &−2 &1 \\ 4 &−1 &2 &0 \end{matrix} \right] \).

\(\left\{ \begin{array} {l} x−y+2z=3 \\ 2x+y−2z=1 \\ 4x−y+2z=0 \end{array} \right.\)

Example \(\PageIndex{6}\)

Write the system of equations that corresponds to the augmented matrix: \(\left[ \begin{matrix} 1 &1 &1 &4 \\ 2 &3 &−1 &8 \\ 1 &1 &−1 &3 \end{matrix} \right] \).

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

Use Row Operations on a Matrix

Once a system of equations is in its augmented matrix form, we will perform operations on the rows that will lead us to the solution.

To solve by elimination, it doesn’t matter which order we place the equations in the system. Similarly, in the matrix we can interchange the rows.

When we solve by elimination, we often multiply one of the equations by a constant. Since each row represents an equation, and we can multiply each side of an equation by a constant, similarly we can multiply each entry in a row by any real number except 0.

In elimination, we often add a multiple of one row to another row. In the matrix we can replace a row with its sum with a multiple of another row.

These actions are called row operations and will help us use the matrix to solve a system of equations.

ROW OPERATIONS

In a matrix, the following operations can be performed on any row and the resulting matrix will be equivalent to the original matrix.

• Interchange any two rows.
• Multiply a row by any real number except 0.
• Add a nonzero multiple of one row to another row.

Performing these operations is easy to do but all the arithmetic can result in a mistake. If we use a system to record the row operation in each step, it is much easier to go back and check our work.

We use capital letters with subscripts to represent each row. We then show the operation to the left of the new matrix. To show interchanging a row:

To multiply row 2 by \(−3\):

To multiply row 2 by \(−3\) and add it to row 1:

Example \(\PageIndex{7}\)

Perform the indicated operations on the augmented matrix:

ⓐ Interchange rows 2 and 3.

ⓑ Multiply row 2 by 5.

ⓒ Multiply row 3 by −2−2 and add to row 1.

\( \left[ \begin{array} {ccc|c} 6 &−5 &2 &3 \\ 2 &1 &−4 &5 \\ 3 &−3 &1 &−1 \end{array} \right] \)

ⓐ We interchange rows 2 and 3.

ⓑ We multiply row 2 by 5.

ⓒ We multiply row 3 by \(−2\) and add to row 1.

Example \(\PageIndex{8}\)

ⓐ Interchange rows 1 and 3.

ⓑ Multiply row 3 by 3.

ⓒ Multiply row 3 by 2 and add to row 2.

\( \left[ \begin{array} {ccc|c} 5 &−2 &-2 &-2 \\ 4 &-1 &−4 &4 \\ -2 &3 &0 &−1 \end{array} \right] \)

ⓐ \( \left[ \begin{matrix} −2 &3 &0 &−2 \\ 4 &−1 &−4 &4 \\ 5 &−2 &−2 &−2 \end{matrix} \right] \)

ⓑ \( \left[ \begin{matrix} −2 &3 &0 &−2 \\ 4 &−1 &−4 &4 \\ 15 &−6 &−6 &−6 \end{matrix} \right] \)

ⓒ \( \left[ \begin{matrix} -2 &3 &0 &2 & \\ 3 &4 &-13 &-16 &-8 \\ 15 &-6 &-6 &-6 & \end{matrix} \right] \)

Example \(\PageIndex{9}\)

ⓐ Interchange rows 1 and 2,

ⓑ Multiply row 1 by 2,

ⓒ Multiply row 2 by 3 and add to row 1.

\( \left[ \begin{array} {ccc|c} 2 &−3 &−2 &−4 \\ 4 &1 &−3 &2 \\ 5 &0 &4 &−1 \end{array} \right] \)

ⓐ \( \left[ \begin{matrix} 4 &1 &−3 &2 \\ 2 &−3 &−2 &−4 \\ 5 &0 &4 &−1 \end{matrix} \right] \) ⓑ \( \left[ \begin{matrix} 8 &2 &−6 &4 \\ 2 &−3 &−2 &−4 \\ 5 &0 &4 &−1 \end{matrix} \right] \) ⓒ \( \left[ \begin{matrix} 14 &−7 &−12 &−8 \\ 2 &−3 &−2 &−4 \\ 5 &0 &4 &−1 \end{matrix} \right] \)

Now that we have practiced the row operations, we will look at an augmented matrix and figure out what operation we will use to reach a goal. This is exactly what we did when we did elimination. We decided what number to multiply a row by in order that a variable would be eliminated when we added the rows together.

Given this system, what would you do to eliminate x ?

This next example essentially does the same thing, but to the matrix.

Example \(\PageIndex{10}\)

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \( \left[ \begin{array} {cc|c} 1 &−1 &2 \\ 4 &−8 &0 \end{array} \right] \)

To make the 4 a 0, we could multiply row 1 by \(−4\) and then add it to row 2.

Example \(\PageIndex{11}\)

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \( \left[ \begin{array} {cc|c} 1 &−1 &2 \\ 3 &−6 &2 \end{array} \right] \)

\( \left[ \begin{matrix} 1 &−1 &2 \\ 0 &−3 &−4 \end{matrix} \right] \)

Example \(\PageIndex{12}\)

Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented matrix: \( \left[ \begin{array} {cc|c} 1 &−1 &3 \\ -2 &−3 &2 \end{array} \right] \)

\( \left[ \begin{matrix} 1 &−1 &3 \\ 0 &−5 &8 \end{matrix} \right] \)

Solve Systems of Equations Using Matrices

To solve a system of equations using matrices, we transform the augmented matrix into a matrix in row-echelon form using row operations. For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

ROW-ECHELON FORM

For a consistent and independent system of equations, its augmented matrix is in row-echelon form when to the left of the vertical line, each entry on the diagonal is a 1 and all entries below the diagonal are zeros.

Once we get the augmented matrix into row-echelon form, we can write the equivalent system of equations and read the value of at least one variable. We then substitute this value in another equation to continue to solve for the other variables. This process is illustrated in the next example.

How to Solve a System of Equations Using a Matrix

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 3x+4y=5 \\ x+2y=1 \end{array} \right. \)

Example \(\PageIndex{14}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 2x+y=7 \\ x−2y=6 \end{array} \right. \)

The solution is \((4,−1)\).

Example \(\PageIndex{15}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 2x+y=−4 \\ x−y=−2 \end{array} \right. \)

The solution is \((−2,0)\).

The steps are summarized here.

SOLVE A SYSTEM OF EQUATIONS USING MATRICES.

• Write the augmented matrix for the system of equations.
• Using row operations get the entry in row 1, column 1 to be 1.
• Using row operations, get zeros in column 1 below the 1.
• Using row operations, get the entry in row 2, column 2 to be 1.
• Continue the process until the matrix is in row-echelon form.
• Write the corresponding system of equations.
• Use substitution to find the remaining variables.
• Write the solution as an ordered pair or triple.
• Check that the solution makes the original equations true.

Here is a visual to show the order for getting the 1’s and 0’s in the proper position for row-echelon form.

We use the same procedure when the system of equations has three equations.

Example \(\PageIndex{16}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 3x+8y+2z=−5 \\ 2x+5y−3z=0 \\ x+2y−2z=−1 \end{array} \right. \)

Example \(\PageIndex{17}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 2x−5y+3z=8 \\ 3x−y+4z=7 \\ x+3y+2z=−3 \end{array} \right. \)

\((6,−1,−3)\)

Example \(\PageIndex{18}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} −3x+y+z=−4 \\ −x+2y−2z=1 \\ 2x−y−z=−1 \end{array} \right. \)

\((5,7,4)\)

So far our work with matrices has only been with systems that are consistent and independent, which means they have exactly one solution. Let’s now look at what happens when we use a matrix for a dependent or inconsistent system.

Example \(\PageIndex{19}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x+y+3z=0 \\ x+3y+5z=0 \\ 2x+4z=1 \end{array} \right. \)

Example \(\PageIndex{20}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x−2y+2z=1 \\ −2x+y−z=2 \\ x−y+z=5 \end{array} \right. \)

no solution

Example \(\PageIndex{21}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} 3x+4y−3z=−2 \\ −2x+3y−z=−1 \\ 2x+y−2z=6 \end{array} \right. \)

The last system was inconsistent and so had no solutions. The next example is dependent and has infinitely many solutions.

Example \(\PageIndex{22}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x−2y+3z=1 \\ x+y−3z=7 \\ 3x−4y+5z=7 \end{array} \right. \)

Example \(\PageIndex{23}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x+y−z=0 \\ 2x+4y−2z=6 \\ 3x+6y−3z=9 \end{array} \right. \)

infinitely many solutions \((x,y,z)\), where \(x=z−3;\space y=3;\space z\) is any real number.

Example \(\PageIndex{24}\)

Solve the system of equations using a matrix: \(\left\{ \begin{array} {l} x−y−z=1 \\ −x+2y−3z=−4 \\ 3x−2y−7z=0 \end{array} \right. \)

infinitely many solutions \((x,y,z)\), where \(x=5z−2;\space y=4z−3;\space z\) is any real number.

Access this online resource for additional instruction and practice with Gaussian Elimination.

• Gaussian Elimination

Key Concepts

• Interchange any two rows
• Multiply a row by any real number except 0
• Add a nonzero multiple of one row to another row