## Linear Equations 3Variables

We will now learn how to solve linear equations in three variables. The system of linear equations is generally in the form

a1x+b1 y+c1z = d1

a2x+b2 y+c2z = d2

a3x+b3 y+c3z = d3

To solve these equations, we select a variable, either x or y or z (usually z is chosen), and eliminate it from the system of equations.

We then obtain simultaneous linear equations in two variables x and y, which we solve as we did in section 2.3. After obtaining the solutions to x and y, we substitute these values in any one of the original equations to obtain the value of z.

Consider these examples.

Example 1

Solve the linear equations.

x+2y+2z=11         ----------- (1)

2x+y+z=7             ------------ (2)

3x+4y+z=14         ------------ (3)

To eliminate z from (1), (2) and (3), we multiply equations 2 and 3 by 2.

2∗ (2x+y+z=7)

2∗ (3x+4y+z=14)

4x+2y+2z=14        ------- (4)

6z+8y+2z=28        ------- (5)

Subtracting 1 from 4 and 1 from 5, we get

4x+ 2y+ 2z=14

-x ± 2y ± 2z=-11

____________

3x= 3

∴    x=3/3

x=1

6x+ 8y + 2z =28

-x ± 2y ± 2z =-11

___________

5x+6y = 17

Substitute x=1 in this equation.

5∗1+6y = 17

6y =17-5

6y =12

y =12/6

y =2

Substitute x=1, y=2 in equation 1.

x+2y+2z = 11

1+2 ∗2+2z = 11

1+4+2z = 11

2z = 11-5

2z = 6

z = 6/2

z = 3

Solution set = {(1,2,3)}.

Example 2

Solve the equations

3x-4y =6z-16

4x -y -z = 5

x=3y+2(z-1)

Rearrange the terms to obtain the general form.

3x-4y-6z = -16 -(1)

4x-y-z = 5    -(2)

x-3y-2z = -2   -(3)

We will eliminate z from the system of equations by multiplying equations (2) and (3) by 6 and 3, respectively. We thus obtain

6∗  (4x-y-z = 5 )

3∗ ( x-3y-2z = -2 )

or

24x-6y-6z = 30 ---------(4)

3x-9y-6z = -6  ----------(5)

Subtracting (1) from (4) we get

24x-6y-6z = 30

-3x 4y 6z = 16

-----------------------

21x-2y = 46 ----------(6)

Subtracting (1) from (5) we get

3x - 9y - 6z = -6

-3x 4y 6z = 16

---------------------------

-5y = 10

y = 10/-5

y = -2

Substituting y = -2 in (6) we get

21x - 2∗(-2) =46

21x + 4 =46

21x = 46-4

21x = 42

x = 42/21

x = 2

Substituting x = 2, y = -2 in equation (1) we get

3∗2-4∗(-2) -6z = -16

6 + 8 - 6z =- 16

14 - 6z = - 16

- 6z = - 16-14

- 6z = -30

z = - 30/-6

z = 5

∴ x = 2, y = -2 z = 5

Solution set = {(2, -2, 5)}

Example 3

Solve the equations.

x-y/5 = 6         (1)

y-z/7 = 8         (2)

z-x/2 = 10       (3)

(1) can be written as

x-y/5 = 6

x = 6+y/5

x= (30+y)/5

Substituting for x in equation (3) we get

z -1/2(30+y)/5 = 10

z-(30+y/10) = 10

(10z-30-y)/10 = 10

10z - 30 - y =10x10

10z - y - 30 =100

10z - y = 100+30

10z-y = 130            (4)

Equation (2) becomes

y -z/7 = 8

(7y-z)/7= 8

7y - z = 8 ∗7

7y - z = 56       ------------ (5)

-y + 10z = 130   ------------(4)

7y - z = 56        ------------ (5)

Multiplying 4 by 7

7(-y + 10z = 130)

7y - z = 56

_______________

69z = 966

z = 966/69

z = 14

Substitute z = 14 into equation (5)

7y - 14 = 56

7y = 56 + 14

7y = 70

y = 70/7

∴       y = 10

Substitute y = 10 in equation (1)

x -10/5 = 6

x - 2 = 6

x = 6+2

x = 8

Solution set ={( 8, 10, 14 )}

Example 4

Solve the equations.

(y+z)/4 = (z+x)/3 = (x+y)/2

x + y + z = 27

We need to reduce these equations to a recognizable form, such as

a1x + b1y + c1 z = d1

a2x + b2y + c2 z = d2

a3x + b3y + c3 z = d3

Consider =(y+z)/4 = (z+x)/3

Cross multiplying 3(y + z) = 4(z + x)

3y + 3z = 4z + 4x

4x + 4z - 3y - 3z = 0

∴            4x - 3y + z = 0               (1)

Consider (z+x)/3 = (x+y)/2

Cross multiplying

2(z + x) = 3(x + y)

2z + 2x = 3x + 3y

3x + 3y - 2z - 2x = 0

x + 3y - 2z = 0         (2)

We now have the following equations

4x - 3y + z = 0      (1)

x - 3y - 2z = 0      (2)

x + y + z = 27    (3)

We eliminate y from these equations. Multiplying equation (3) with 3 we get

3∗ (x + y + z = 27)

3x + 3y + 3z = 81         (4)

4x - 3y + z = 0

3x + 3y + 3z = 81

______________

7x +4z = 81           (5)

Adding (1) and (2) we get

4x - 3y + z = 0

x + 3y -2z = 0

_____________

5x- z = 0

5x = z

Substituting z = 5x in (5) we get

7x + 4 ∗5x = 81

7x + 20x = 81

27x = 81

x = 81/27

x = 3

z = 5x

z = 5∗3

z = 15

Substitute x = 3, z = 15 in equation (1)

4x - 3y + z = 0

4∗3 - 3y +15 = 0

12 - 3y + 15 = 0

- 3y + 27 = 0

-3y = -27

y = -27/-3

y = 9

Solution set = {( 3, 9, 15 )}

#### Try these questions

Solve the following equations

1. x + 3y + 4z = 14
x + 2y + z = 7
2x + y + 2z = 2
Let x + 3y + 4z =14     (1)
x + 2y + z = 7     (2)
2x + y + 2z = 2    (3)
To eliminate z from the equations multiply (2) by 4 and (3) by 2.
4∗ (x + 2y + z = 7)
2∗ (2x +y +2z = 2)
4x + 8y + 4z = 28       (4)
4x + 2y + 4z = 4         (5)
Subtracting (5) from (4) Subtracting (1) from (4) we get
4x + 8y + 4z = 28
- x ± 3y ± 4z = - 14
-----------------------------
3x + 5y = 14
Substituting y = 4 in this equation
3x + 5∗4 =14
3x + 20 = 14
3x = 14-20
3x =- 6
x = -6/3
x = -2
Substituting x = -2 y = 4 in equation (3)
2∗(-2) + 4 + 2z = 2
- + + 2z = 2
2z = 2
z = 2/2
z = 1
Solution set = { (-2, 4, 1) }

2. x + 4y + 3z = 17
3x + 3y + z = 16
2x + 2y + z = 11
Answer:  x + 4y + 3z = 17      (1)
3x + 3y + z = 16      (2)
2x + 2y + z = 11      (3)
To eliminate z from the equations multiply (2) and (3) by 3.
3∗ ( 3x+ 3y + z = 16)
3∗ (2x + 2y + z = 11)
9x + 9y + 3z = 48       (4)
6x + 6y + 3z = 33       (5)
Subtracting (5) from (4) Subtracting (1) from (5) we get
6x + 6y + 3z = 33
- x ± 4y ± 3z = -17
---------------------------
5x + 2y = 16             (7)
To eliminate y multiply (6) by 2 and (7) by 3 and subtract
2 (3x + 3y = 15)
3 (5x + 2y = 16) x = -18/-9
x = 2
Substituting x = 2 in (6)
3∗2 + 3y = 15
3y = 15-6
3y = 9
y = 9/3
y = 3
Substituting x = 2, y = 3 in equation (1)
2 + 4 ∗3 + 3z = 17
2 + 12 + 3z = 17
3z = 17 - 14
3z = 3
z = 3/3
z = 1
Solution set = { (2, 3, 1) }

3. x - 2y + 3z = 2
2x - 3y + z = 1
3x - y + 2z = 9 Answer:  x - 2y + 3z = 2        (1)
2x - 3y + z = 1        (2)
3x - y + 2z = 9        (3)
To eliminate y multiply (1) by 2 and again by 3
2(x - 2y + 3z = 2)
2x - 4y + 6z = 4       (4)
3(x - 2y + 3z = 2)
3x - 6y + 9z = 6)       (5)
Subtracting (2) from (4) Subtracting 3 from 5 To eliminate y multiply (6) by (5) and subtract (7) from the result
5(-y + 5z = 3) Substituting z = 1 in (6)
-y + 5 ∗1 = 3
-y = 3-5
-y = -2
y = 2
Substituting y = 2, z = 1 in (1)
x - 2 ∗2 + 3 ∗1 = 2
x - 4 + 3 =2
x - 1 = 2
x = 2+1
x = 3
Solution set = { (3, 2, 1) }

4. 5x + 2y = 14
y - 6z = -15
x + 2y+z = 0
Answer:   5x + 2y = 14          (1)
y - 6z = -15          (2)
x + 2y + z = 0      (3)
consider (2)
y - 6z = -15
y = -15 + 6z
Substituting y = -15+6z in (1) and (3) we get
5x + 2 (-15+6z) = 14
5x - 30 + 12z = 14
5x + 12z = 14+30
5x + 12z = 44             (4)
x+2 (-15+6z) + z = 0
x - 30 + 12z + z = 0
x + 13z = 30          (5)
To eliminate x from the equations. Multiply (5) by 5 and subtracting (4) from the result we get
5(x + 13z = 30)
5x + 65z = 150
-5x ± 12z = -44
------------------------
53z = 106
z = 106/53
z = 2
Substituting z = 2 in (5)
x + 13∗2 = 30
x + 26 = 30
x = 30 - 26
x = 4
Substituting z = 2 in (2)
y - 6 ∗2 = -15
y = -15+12
y = -3
Solution set = { (4, -3, 2) }

5.  y-z / 3 = y-x / 2= 5z - 4x
y + z = 2x + 1
Answer: Reducing these equations to the general from of
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Consider
y-z/3 = y-x/2
Cross multiplying
2(y-z) = 3(y-x)
2y-2z = 3y -3x
2y - 2z - 3y+3x = 0
3x - y - 2z = 0
Consider
y-x /2 = 5z-4x / 1
y - x = 2(5z - 4x)
y - x = 10z - 8x
y - x - 10z + 8x = 0
7x + y -10z = 0
Consider
y + z = 2x + 1
2x+1 - y - z = 0
2x - y - z = -1
We now have the following equations.
3x - y - 2z = 0             (1)
7x + y - 10z = 0             (2)
2x - y-z = -1            (3)
We can eliminate y from these equations
Adding (1) and (2) we get Adding (2) and (3) we get To eliminate x from (4) and (5) multiply (4) by 9 and (5) by 10 and subtracting
9(10x - 12z = 0)
10(9x - 11z = -1) -------------------------
2z = 10
z = 10/2
z = 5
Substituting z = 5 in (4)
10x - 12 ∗5 = 0
10x - 60 = 0
10x = 60
x = 60/10
x = 6
Substituting x = 6, z = 5 in (3)
2∗6 - y - 5 = -1
12 - y - 5 = -1
7-y = -1
- y = -1 -7
-y = -8
y = -8/-1
y = 8
Solution set = { (6, 8, 5) }