Linear Equations with Three Variables |
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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. |
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Example 1 |
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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
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3x= 3
∴ x=3/3
x=1
6x+ 8y + 2z =28
-x ± 2y ± 2z =-11
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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)}. |
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Example 2 |
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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
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21x-2y = 46 ----------(6)
Subtracting (1) from (5) we get
3x - 9y - 6z = -6
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-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)} |
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Example 3 |
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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)
Adding -7y +70z = 910
7y - z = 56
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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 )} |
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Example 4 |
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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)
Adding (1) and (4)
4x - 3y + z = 0
3x + 3y + 3z = 81
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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 )} |
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Try these questions |
Solve the following equations |
1. |
x + 3y + 4z = 14
x + 2y + z = 7
2x + y + 2z = 2 |
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Answer |
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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) |
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Subtracting (1) from (4) we get
4x + 8y + 4z = 28
- x ± 3y ± 4z = - 14
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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
z = 2/2
z = 1
Solution set = { (-2, 4, 1) } |
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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)
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Subtracting (1) from (5) we get
6x + 6y + 3z = 33
- x ± 4y ± 3z = -17
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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) |
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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) } |
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3. |
x - 2y + 3z = 2
2x - 3y + z = 1
3x - y + 2z = 9 |
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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)
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Subtracting 3 from 5 |
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To eliminate y multiply (6) by (5) and subtract (7) from the result
5(-y + 5z = 3) |
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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) } |
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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
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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) } |
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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)
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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) } |
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