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How To Solve Three Variable Linear Equations

By Kathleen Knowles, 07 Oct 2020

An equation between two variables is known as a Linear equation. A Linear equation usually results in a straight line when plotted graphically. The variables, which on most occasions are real numbers, are considered the parameters of the equation.

There are several systems of linear equations involving the same set of variables. A system here refers to when you have two or more equations working together.

Examples Relating to Three Variable Linear Equations

2X + Y=6

3X - Y= 4

X-2Y +3Z=9

-X+3Y-Z=-6

X+2Y+3Z=-7

2X-3Y-5Z=9

-6X-8Y+Z=-22

Using the Elimination Method to Solve a Three Variable Linear Equation

A three-variable linear equation is a bit more difficult to solve compared to equations with two variables. This complexity is a result of the additional variable.

Although there are several methods for solving this type of equation, the elimination method remains the most straightforward.

Follow the procedure below to use the elimination method in solving three variable linear equations:

1. Put all the equations in standard form, avoiding decimals and fractions.

2. Select a suitable variable to eliminate.

3. Choose any two of the three equations and eliminate the selected variable.

4. Select a different set of the two equations and eliminate the same selected variable.

5. Solve the two equations in steps three and four for the two variables they contain.

6. Substitute the answers in step five into any equation that has the remaining variable.

7. Check to prove the solution with all the three original equations.

Example 1

Task: Solve the following system of equations using the elimination method.

X-2Y +3Z=9 ______________(1)

-X+3Y-Z=-6_______________ (2)

2X-5Y+5Z=17_____________ (3)

Solution

In this case, you realize that the system is already in standard form; therefore, choose the variable to eliminate, for instance, X. You should then select two equations with which to eliminate X. For this example, pick (1) and (2).

Thus;

X-2Y +3Z=9

-X+3Y-Z=-6

__________________

Y +2Z=3__________(4)

You should then select a different set of two equations, in this case, equations (1) and (3) to eliminate the same variable.

Thus;

X-2Y+3Z=9___________(1)

2X-5Y +5Z=17_________(3)

To eliminate X, multiply equation (1) with (-2) or any other suitable constant.

Thus;

-2X +4Y -6Z = -18

2X-5Y+5Z=17

____________________

-Y-Z=-1__________(5)

You can now solve equation (4) and (5) .

Y+2Z=3

-Y-Z=-1

__________________

Z=2

Now substitute Z=2 in any of the created 2×2 system, that is (4) or (5). In this case pick equation (5).

-Y-Z=-1

-Y-2=-1

Thus Y=-1

Use the answers obtained to substitute into any equation involving the remaining variable. In this case pick X-2Y+3Z=9___________(1)

X-2Y+3Z=9___________(1)

X-2(-1)+3(2)=9

X+2+6=9

X+8=9

X=1

Therefore the solution is X=1, Y =-1 and Z = 2.

Example 2

Task: Solve the following system of equations using the elimination method.

X=3Z-5

2X+2Z=Y +16

7X-5Z= 3Y +19

Solution

First, write all the equations in standard form.

Thus;

X -3Z = -5____________(1)

2X-Y+2Z=16_________(2)

7X-3Y-5Z=19_________(3)

Next, choose a suitable variable to eliminate. In this case, select Y since it is missing in equation (1). Then, select equation (2) and (3) to eliminate Y.

2X-Y +2Z=16__________(2)

7X-3Y -5Z=19__________(3)

Please, multiply equation (2) by a suitable constant,say ( -3) to get:

-6X+3Y-6Z= 48

Then, eliminate Y.

-6X+3Y-6Z= 48

7X-3Y -5Z=19

_____________________

X-11Z=-29_________________(4)

It would be best if you then solved equation (1) and (4) since they are 2×2 system linear equation by subtracting.

Thus;

X -3Z = -5_________________(1)

X-11Z=-29_________________(4)

______________________

8Z=24

Therefore Z=3

You can then substitute Z=3 into equation (4)

X-11Z=-29_________________(4)

X-11(3)=-29

X-33=-29

X=4

Then, submit X=4 and Z=3 into equation 3 to get the value of Y.

7X-3Y -5Z=19__________(3)

7(4)-3Y-5(3)=19

28-3Y-15=19

-3Y=6

Therefore Y=-2

Here's the solution: X=4,Y=-2 and Z=3

Consider the Gaussian Elimination Method in Solving Three Variable Linear Equations.

The Gaussian Elimination Method is the best method for solving three (or more) variable equations. However, the Gaussian Elimination Method is generally for experts, as it involves a bit of set up work. You don't have to worry though, because once you have the system set up, the equation is straightforward to solve.

To use the Gaussian Elimination Method, you have to translate your equations into a matrix using placeholder zeros and ones, where necessary. You should then set up the matrix by manipulating it through basic application of the properties of matrixes. You should then add, subtract, multiply and divide the rows in the matrix to make ones and zeros in specific locations.

Examples

Solve the following equation using the Gaussian Elimination Method

system of equations.

–3x + 2y – 6z = 6

5x + 7y – 5z = 6

x + 4y – 2z =8

Solution

You realize that in this problem, no expression for a variable is solved. It means you'll have to do the multiplication and addition tasks to simplify the system.

First, do away with the leading x terms in any of the two rows. To begin, simply pick any two rows you find easy to clear out before switching the rows later to have the system in the upper triangular form.

Since there is no existing rule limiting you to pick the x term starting from the first row, pick the simpler x term in the third row which gives a simple coefficient of multiply the third row by 3 then add it to the first row. If you find this difficult, you can do the calculations on paper.

Thus;

-3X + 2Y-6Z=6

5X + 12Y-6Z=24

14Y - 12Z=30

You should now write the results in this format:

7Y- 6Z= 15

5X +7Y- 5Z= 6

X + 4Y -2Z= 8

Whereas in solving two variable expressions, you are allowed to multiply a row, there is no provision for this in three variable linear equations. It is therefore suggested that you work it out on a scratch paper. You should get this result:

14Y -12Z= 30

5X +7Y - 5Z= 6

X + 4Y - 2Z= 8

Next, multiply the first row by one- half to get smaller values for coefficients.

Thus;

7Y - 6Z = 15

5X + 7Y -5Z= 6

X +4Y - 2Z= 8

Now that you have the value of Z, divide the first row by 43. Next, arrange the rows in the upper triangular format:

Thus;

43Z = 43

Y- 7Z= -4

X +4Y- 2Z= 8

Multiply the first row by 1/43 to get:

Z= 1

Y-7Z=- 4

X+ 4Y-2Z= 8

 

X+ 4Y-2Z=8

Y-7Z=-4

Z=1

Please use the back solving process to get the values of Y and Z:

Y – 7(1) = –4

Y – 7 = –4

Y = 3

 

X + 4(3) – 2(1) = 8

X + 12 – 2 = 8

X + 10 = 8

X = –2

Thus the solution is:

X = - 2

Y= 3

Z = 1

Try Other Methods of Solving Three Variable Linear Equation.

The remaining methods of solving three variable linear equations are the substitution method and the graphical method. In substitution, the equation can quickly be written with a single variable, as the subject is identified. The identification is made by solving the equation of the variable. Next, substitution is done to the expression in which the variable appears as well as the two other equations. You will, therefore, come up with a system with fewer variables. After you have solved the smaller system, either through substitution or any other method, substitute the solutions you have found for the variables back into the first right-hand side of the expression.

Consider this example:

Solve the following three variable linear equations by the substitution method.

3X +2Y-Z= 6

-2X+2Y +Z=3

X+Y+2=4

The coefficient of Z is already Z in equation (1). Therefore, solve for Z to get:

Z= 3X+2Y-6

You can now substitute for Z in equation (2) and (3).

Thus;

-2X + 2Y +(3X+2Y-6)=3

X+Y+(3X+2Y-6)=4

You can then simplify the new system to:

X+4Y=9

4X+3Y=10

Now solve for Z in equation ( 1)

X=9-4Y

Substitute the expression X=9-4Y in equation (3).

Thus;

4(9-4Y)+3Y=10

36-16Y+3Y=10

13Y=26

Y=3

With Y=2, work back the equation X=9-4Y.

Thus,

X=9-4Y

X=1

You can now find Z's value by substituting X=1 and Y= 2 in equation (2).

Z=3X +2Y-6

Z=3(1)+2(1)-6

Z=1.

Therefore X= 1,Y=2 and Z=1.

Conclusion

Knowledge of three variable linear equations has actual real-life applications. In tackling life's challenges, many situations can be converted into mathematical expressions to illustrate the relationship between variables, otherwise known as linear equations.

Finally, knowledge of three variable equations is useful in supply and demand economics.

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