System of Three Variable Linear Equations: Features, Components, Solving Methods and Example Problems

System of Three Variable Linear Equations: Features, Components, Solving Methods and Example Problems –  What is meant by a system of three variable equations? At this opportunity About the knowledge.co.id will discuss it and of course also the things that surround it. Let's look at the discussion together in the article below to better understand it.

System of Three Variable Linear Equations: Features, Components, Solving Methods and Example Problems


The system of three-variable equations or commonly abbreviated as SPLTV is a collection of linear equations that have three variables. A linear equation is characterized by the highest exponential of the variables in the equation being one. In addition, the sign connecting the equations is an equals sign.

In architecture, there are mathematical calculations for constructing buildings, one of which is a system of linear equations. A system of linear equations is useful for determining the coordinates of intersection points. Precise coordinates are essential to produce a building that fits the sketch. In this article, we will discuss a system of three variable linear equations (SPLTV).

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System of Three Variable Linear Equations - is an extended form of a system of two variable linear equations (SPLDV). Which, in a three-variable system of linear equations consists of three equations, each equation has three variables (eg x, y and z).

The system of three-variable linear equations consists of several linear equations with three variables. The general form of the three-variable linear equation is as follows.

ax + by + cz = d

a, b, c, and d are real numbers, but a, b, and c cannot all be 0. This equation has many solutions. One solution can be obtained by comparing arbitrary values ​​to two variables to determine the value of the third variable.


Characteristics of a Three Variable Linear Equation System

An equation is called a three-variable system of linear equations if it has the following characteristics:

  • Using an equals sign (=) relation
  • Has three variables
  • The three variables have degree one (rank one)

Three Variable Linear Equation System Components

Contains three components or elements that are always related to a three-variable system of linear equations.

The three components are: terms, variables, coefficients and constants. The following is an explanation of each of the SPLTV components.

  • Ethnic group

Term is a part of an algebraic form consisting of variables, coefficients and constants. Each term is separated by adding or subtracting punctuation marks.

Example:

6x – y + 4z + 7 = 0, then the terms of the equation are 6x, -y, 4z and 7.

  • Variable

Variables are variables or substitutes for a number which are generally denoted by the use of letters such as x, y and z.

Example:

Yulisa has 2 apples, 5 mangoes and 6 oranges. If we write in the form of an equation then:

For example: apples = x, mangoes = y and oranges = z, so the equation is 2x + 5y + 6z.

  • Coefficient

The coefficient is a number that expresses the number of variables of the same type.

The coefficient is also known as the number in front of the variable, because the writing of an equation for the coefficient is in front of the variable.

Example:

Gilang has 2 apples, 5 mangoes and 6 oranges. If we write it in the form of an equation then:

For example: apples = x, mangoes = y and oranges = z, so the equation is 2x + 5y + 6z.

From this equation, it can be seen that 2, 5 and 6 are coefficients where 2 is the x coefficient, 5 is the y coefficient and 6 is the z coefficient.

  • Constant

A constant is a number that is not followed by a variable, so it will have a fixed or constant value regardless of the value of the variable or variables.

Example:

2x + 5y + 6z + 7 = 0, from this equation the constant is 7. This is because 7 has a fixed value and is not affected by any variables.


Method of Solving a System of Three Variable Linear Equations

A value (x, y, z) is a set of solutions to a system of three-variable linear equations if the value (x, y, z) satisfies the three equations in SPLTV. The set of SPLTV solutions can be determined in two ways, namely the method of substitution and method of elimination.

  • Substitution Method

The substitution method is a method for solving systems of linear equations by substituting the value of one of the variables from one equation to another. This method is carried out until all variable values ​​are obtained in a three-variable system of linear equations.

The substitution method is easier to use on SPLTV which contains an equation with a coefficient of 0 or 1. The following are the steps for solving with the substitution method.

  1. Find an equation that has a simple form. Equations with simple forms have coefficients of 1 or 0.
  2. Express one of the variables in the form of two other variables. For example, the variable x is expressed in terms of the variable y or z.
  3. Substitute the variable values ​​obtained in the second step into the other equations in SPLTV, so that a two-variable linear equation system (SPLDV) is obtained.
  4. Determine the SPLDV solution obtained in step three.
  5. Determine the values ​​of all unknown variables.

Let's try to do the following example problem. Determine the set of solutions to the three-variable system of linear equations below.

x + y + z = -6 … (1)

x – 2y + z = 3 … (2)

-2x + y + z = 9 … (3)

First, we can change equation (1) to, z = -x – y – 6 to equation (4). Then, we can substitute equation (4) into equation (2) as follows.

x – 2y + z = 3

x – 2y + (-x – y – 6) = 3

x – 2y – x – y – 6 = 3

-3y = 9

y = -3

After that, we can substitute equation (4) into equation (3) as follows.

-2x + y + (-x – y – 6) = 9

-2x + y – x – y – 6 = 9

-3x = 15

x = -5

We've got the values ​​x = -5 and y = -3. We can plug it into equation (4) to get the value of z as follows.

z = -x – y – 6

z = -(-5) – (-3) – 6

z = 5 + 3 – 6

z = 2

So, we get the solution set (x, y, z) = (-5, -3, 2)

  • Elimination Method

The elimination method is a method of solving a system of linear equations by eliminating one of the variables in two equations. This method is carried out until there is only one variable left.

The elimination method can be used for all three-variable systems of linear equations. But this method requires long steps because each step can only eliminate one variable. A minimum of 3 times the elimination method is required to determine the set of SPLTV solutions. This method is easier when combined with the substitution method.

The steps for solving using the elimination method are as follows.

  1. Observe the three equations on SPLTV. If there are two equations that have the same coefficient value on the same variable, subtract or add up the two equations so that the variable has a coefficient of 0.
  2. If neither variable has the same coefficient, multiply both equations by the number that makes the coefficient of a variable in both equations the same. Subtract or add up the two equations so that the variable has a coefficient of 0.
  3. Repeat step 2 for the other pair of equations. The variables omitted in this step must be the same as the omitted variables in step 2.
  4. After obtaining two new equations in the previous step, determine the set of solutions for the two equations using the two-variable system of linear equations (SPLDV) solution method.
  5. Substitute the values ​​of the two variables obtained in step 4 in one of the SPLTV equations to obtain the value of the third variable.

We will try to use the elimination method in the following questions. Determine the set of SPLTV solutions!

2x + 3y – z = 20 … (1)

3x + 2y + z = 20 … (2)

X + 4y + 2z = 15 … (3)

SPLTV can be determined the set of solutions by eliminating the variable z. First, add up equations (1) and (2) to obtain:

2x + 3y – z = 20

3x + 2y + z = 20 +

5x + 5y = 40

x + y = 8 … (4)

Then, multiply 2 in equation (2) and multiply 1 in equation (1) to get:

3x + 2y + z = 20 |x2 6x + 4y + 2z = 40

x + 4y + 2z = 15 |x1 x + 4y + 2z = 15

5x = 25

x = 5

After knowing the value of x, substitute it into equation (4) as follows.

x + y = 8

5 + y = 8

y = 3

Substitute the x and y values ​​in equation (2) as follows.

3x + 2y + z = 20

3(5) + 2(3) + z = 20

15 + 6 + z = 20

z = -1

So that the set of solutions for SPLTV (x, y, z) is (5, 3, -1).

  • Combined or Mixed Methods

Solving for systems of linear equations using combined or mixed methods is a way of solving by combining two methods at once.

The method in question is the elimination method and the substitution method.

This method can be used by using the substitution method first or by elimination first.

And this time, we will try a combined or mixed method with 2 techniques, namely:

Eliminate first and then use the substitution method.
Substituting first and then using the elimination method.

The process is almost the same as in solving SPLTV using the elimination method and the substitution method.

In order for you to understand more about how to solve SPLTV using this combination or mixture, here we provide some examples of questions and their discussion.


Problems example


Problem 1.

Determine the set of SPLTV solutions below using the substitution method:
x – 2y + z = 6
3x + y – 2z = 4
7x – 6y – z = 10

Answer:

The first step is to first determine the simplest equation.

Of the three equations, the first equation is the simplest. From the first equation, express the variables x as a function of y and z as follows:

⇒ x – 2y + z = 6

⇒ x = 2y – z + 6

Substitute the variable or variables x into the second equation

⇒ 3x + y – 2z = 4

⇒ 3(2y – z + 6) + y – 2z = 4

⇒ 6y – 3z + 18 + y – 2z = 4

⇒ 7y – 5z + 18 = 4

⇒ 7y – 5z = 4 – 18

⇒ 7y – 5z = –14 …………… Eq. (1)

Substitute the variable x into the third equation

⇒ 7x – 6y – z = 10

⇒ 7(2y – z + 6) – 6y – z = 10

⇒ 14y – 7z + 42 – 6y – z = 10

⇒ 8y – 8z + 42 = 10

⇒ 8y – 8z = 10 – 42

⇒ 8y – 8z = –32

⇒ y – z = –4 ……………… Eq. (2)

Equations (1) and (2) form SPLDV y and z:
7y – 5z = –14
y – z = –4

Then solve the SPLDV above using the substitution method. Choose one of the simplest equations. In this case the second equation is the simplest equation.

From the second equation, we get:

⇒ y – z = –4

⇒ y = z – 4

Substitute the variable y into the first equation

⇒ 7y – 5z = –14

⇒ 7(z – 4) – 5z = –14

⇒ 7z – 28 – 5z = –14

⇒ 2z = –14 + 28

⇒ 2z = 14

⇒ z = 14/2
⇒ z = 7

Substitute the value z = 7 into one of the SPLDV, for example y – z = –4 so we get:

⇒ y – z = –4

⇒ y – 7 = –4

⇒ y = –4 + 7

⇒ y = 3

Then, substitute the values ​​y = 3 and z = 7 to one of the SPLTV, for example x – 2y + z = 6 so we will get:

⇒ x – 2y + z = 6

⇒ x – 2(3) + 7 = 6

⇒ x – 6 + 7 = 6

⇒ x + 1 = 6

⇒ x = 6 – 1

⇒ x = 5

Thus, we get x = 5, y = 3 and z = 7. So that the set of solutions for the SPLTV problem is {(5, 3, 7)}.
In order to ensure that the x, y, and z values ​​obtained are correct, we can find out by substituting the x, y, and z values ​​into the three SPLTV above. Among others:

Equation I:

⇒ x – 2y + z = 6

⇒ 5 – 2(3) + 7 = 6

⇒ 5 – 6 + 7 = 6

⇒ 6 = 6 (true)

Equation II:

⇒ 3x + y – 2z = 4

⇒ 3(5) + 3 – 2(7) = 4

⇒ 15 + 3 – 14 = 4

⇒ 4 = 4 (true)

Equation III:

⇒ 7x – 6y – z = 10

⇒ 7(5) – 6(3) – 7 = 10

⇒ 35 – 18 – 7 = 10

⇒ 10 = 10 (true)
From the data above, it can be ascertained that the x, y and z values ​​that we get are correct and fulfill the system of linear equations of the three variables in question.

Problem 2.

Given a system of linear equations:

(i) x -3y +z =8

(ii) 2x =3y-z =1

(iii) 3x -2y -2z =7

The x+y+z value is

A. -1

B. 2

C. 3

D. 4

Discussion:

From equation (i) x – 3y + z = 8 → x = 3y – z + 8 …. (iv)

Substitute equation (iv) into equation (ii):
2x + 3y – z = 1
2(3y – z + 8) + 3y – z = 1
6y – 2z + 16 + 3y – z = 1
9y – 3z + 16 = 1
3z = 9y + 15
z = 3y + 5 …. (v)

Substitute equation (iv) into equation (iii):
3x – 2y – 2z = 7
3(3y – z + 8) – 2y – 2z = 7
9y – 3z + 24 – 2y – 2z = 7
7y – 5z + 24 = 7
5z = 7y + 24 – 7
5z = 7y + 17…. (vi)

Substitute equation (v) into equation (vi):
5z = 7y + 17
5(3y + 5) = 7y + 17
15y + 25 = 7y + 17
15y – 7y = -25 + 17
8y = -8 → y = –1 …. (vii)

Substitute the value of y = – 1 in equation (vi) to get the z value.
5z = 7y + 17
5z = 7( – 1) + 17
5z = – 7 + 17
5z = 10 → z = 2 … (viii)

Substitute the value y = – 1 and z = 2 in equation (i) to get the value x.
x – 3y + z = 8
x – 3(- 1) + 2 = 8
x + 3 + 2 = 8
x + 5 = 8
x = 8 – 5 → x = 3

The values ​​of the three variables that satisfy the system of equations are obtained, namely x = 3, y = – 1, and z = 2.

So, the value of x + y + z = 3 + (-1) + 2 = 4.

Answer: D

Given a System of Linear Equations

(i) = x – 3y +

Discussion:

From equation (i) x – 3y + z = 8 → x = 3y – z + 8 …. (iv)

Substitute equation (iv) into equation (ii):
2x + 3y – z = 1
2(3y – z + 8) + 3y – z = 1
6y – 2z + 16 + 3y – z = 1
9y – 3z + 16 = 1
3z = 9y + 15
z = 3y + 5 …. (v)

Substitute equation (iv) into equation (iii):
3x – 2y – 2z = 7
3(3y – z + 8) – 2y – 2z = 7
9y – 3z + 24 – 2y – 2z = 7
7y – 5z + 24 = 7
5z = 7y + 24 – 7
5z = 7y + 17…. (vi)

Substitute equation (v) into equation (vi):
5z = 7y + 17
5(3y + 5) = 7y + 17
15y + 25 = 7y + 17
15y – 7y = -25 + 17
8y = -8 → y = – 1 …. (vii)

Substitute the value of y = – 1 in equation (vi) to get the z value.
5z = 7y + 17
5z = 7( – 1) + 17
5z = – 7 + 17
5z = 10 → z = 2 … (viii)

Substitute the value y = – 1 and z = 2 in equation (i) to get the value x.
x – 3y + z = 8
x – 3(- 1) + 2 = 8
x + 3 + 2 = 8
x + 5 = 8
x = 8 – 5 → x = 3

The values ​​of the three variables that satisfy the system of equations are obtained, namely x = 3, y = – 1, and z = 2.

So, the value of x + y + z = 3 + (-1) + 2 = 4.

Answer: D

System of Three Variable Linear Equations: Features, Components, Solving Methods and Example Problems

Problem 3.

Determine the solution set of the three-variable system of linear equations below using the combined method.
x + 3y + 2z = 16
2x + 4y – 2z = 12
x + y + 4z = 20

Answer:

Substitution Method (SPLTV)

The first step determines the simplest equation. From the three equations above, we can see that the third equation is the simplest equation.

From the third equation, express the variable z as a function of y and z as follows:

⇒ x + y + 4z = 20

⇒ x = 20 – y – 4z ………… Eq. (1)

Then, substitute equation (1) above into the first SPLTV.

⇒ x + 3y + 2z = 16

⇒ (20 – y – 4z) + 3y + 2z = 16

⇒ 2y – 2z + 20 = 16

⇒ 2y – 2z = 16 – 20

⇒ 2y – 2z = –4

⇒ y – z = –2 …………. Pers. (2)

Then, substitute equation (1) above into the second SPLTV.

⇒ 2x + 4y – 2z = 12

⇒ 2(20 – y – 4z) + 4y – 2z = 12

⇒ 40 – 2y – 8z + 4y – 2z = 12

⇒ 2y – 10z + 40 = 12

⇒ 2y – 10z = 12 – 40

⇒ 2y – 10z = –28 ………… Eq. (3)

From equation (2) and equation (3) we get the SPLDV y and z as follows:
y – z = –2
2y – 10z = –28

Elimination Method (SPLDV)

To eliminate or eliminate y, then multiply the first SPLDV by 2 so that the y coefficients of the two equations are the same.

Next, we differentiate the two equations so that we get z values ​​like the following:

y – z = -2 |×2| → 2y – 2z = -4

2y – 10z = -28 |×1| → 2y – 10z = -28
__________ –
8z = 24
z = 3

To eliminate z, then multiply the first SPLDV by 10 so that the z coefficients in both equations are the same.

Then we subtract the two equations so we will get the y value as follows:

y – z = -2 |×10| → 10y – 10z = -20

2y – 10z = -28 |×1| → 2y – 10z = -28
__________ –
8y = 8
z = 1

Up to this point, we get the values ​​y = 1 and z = 3.

The final step is to determine the value of x. The way to determine the x value is by entering the y and z values ​​into one of the SPLTV. For example x + 3y + 2z = 16 so we will get:

⇒ x + 3y + 2z = 16

⇒ x + 3(1) + 2(3) = 16

⇒ x + 3 + 6 = 16

⇒ x + 9 = 16

⇒ x = 16 – 9

⇒x = 7

That way, we get the values ​​x = 7, y = 1 and z = 3 so that the set of SPLTV solutions for the above problem is {(7, 1, 3)}.

Thus the review from About the knowledge.co.id aboutSystem of Three Variable Linear Equations, hopefully can add to your insight and knowledge. Thank you for visiting and don't forget to read other articles

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