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High School Mathematics
1.23 Linear Equation - Three Variables

Example:1
Solve the following linear equations in three variables:
2x + y -z = 4 -------(1)
x + y + z = 10 -------(2)
2z = 4 -------(3)

Solution:
From equation 3 we have 2z = 4 then z = 4/2 = 2.
Therefore z = 2.
substitute z = 2 in equation 1 and equation 2.
i.e 2x + y -z = 4 => 2x + y - 2 = 4 => 2x + y = 6.
and x + y + z = 10 => x + y + 2 = 10 =>x + y = 8
Solve these equations
2x + y = 6 and x + y = 8

2x + y = 6
-x - y = -8
--------------
x = -2

i.e x = -2.
Substituting values x = -2 and z = 2 in equation 2.
i.e x + y + z = 10
=> -2 + y + 2 = 10
i.e y = 10
Therefore x = -2, y = 10 and z = 2

Verification:
If we substitute x = -2, y = 10 and z = 2 in 2x + y - z = 4, then we have
2x + y - z
= 2(-2) + 10 - 2
= -4 + 10 - 2
= -6 + 10
= 4
Hence verified.


Example:
Solve the following linear equations in three variables.
a + b = 12 -------(1)
b + c = 6-------(2)
c + a = 8-------(3)

Solution:
From equation 1 we have,
a + b = 12 => b = 12 - a

substitute b = 12 - a in equation 2
12 - a + c = 6 => - a + c = - 6 => - a + c = - 6.
Also equation 3 is c + a = 8.
Add both equations.
- a + c = - 6 ------(2)
a + c = 8-------(3)
----------------
2c = 2
therefore c = 1

substituting c = 1 in equation a + c = 8 we get
a + c = 8 => a + 1 = 8 => a = 7.

substituting c = 1 and a = 7 in equation 1(2 or 3) we get
a + b = 12 => 7 + b = 12 => b = 5.

Therefore a = 7, b = 5 and c = 1.

Verification:
If we substitute a = 7, b = 5 and c = 1 in a + b = 12 and b + c = 6, then we have
a + b = 7 + 5 = 12
and b + c = 5 + 1 = 6.
Hence verified.


Example:
Solve the following linear equations in three variables:
a + b + c = 10 --------(1)
a + 2b + 3c = 5 ------(2)
2a + b + 5c = 4-------(3)

Solution:
a + b + c = 10--------equation 1
a = 10 - b - c

a + 2b + 3c = 5 ------equation 2
(10 - b - c) + 2b + 3c = 5 => 10 - b - c + 2b + 3c = 5 => 10 + b + 2c = 5 => b + 2c = -5

2a + b + 5c = 4---------equation 3
2(10 - b - c) + b + 5c = 4 =>20 - 2b - 2c + b + 5c = 4 => 20 - b + 3c = 4 => -b + 3c = -16
b + 2c = -5
-b + 3c = -16 solving these equations we will find b and c.

b + 2c = -5
-b + 3c = -16
-----------------------
0 + 5c = 21
Hence c = 21/5 substituting in 1 we get a + b = -1/5


Directions: Solve the following equations with three variables. Also write at least 5 examples of your own.
Q 1: Solve the following linear equations.
x - 3y + z = 1
2x - y - 2z = 2
x + 2y - 3z = -1

x = 1, y = -1 and z = 3
x = 1, y = -1 and z = -2
inconsistent system
x = -1, y = -1 and z = 2

Q 2: Solve the following linear equations.
x - 2y + 3z = 9
y + 3z = 5
z = 2

x = -1, y = -2 and z = 2
inconsistent system
x = 1, y = -3 and z = 2
x = 1, y = -1 and z = 2

Q 3: Solve the following linear equations.
4x + 3y - z = -20
-6x = 36
4x - 3y -z = -10

x = 4, y = -1/3 and z = -9
inconsistent system
x = -6, y = -4/3 and z = 5
x = -6, y = -5/3 and z = -9

Q 4: Solve the following linear equations.
x - 2y + 3z = 9
-x + 3z = -4
2x - 5y + 5z = 17

inconsistent system
x = 53/8, y = 1/8 and z = 7/8
x = -1, y = 1 and z = 2
x = 51/3, y = -1/8 and z = -5/8

Q 5: Solve the following linear equations.
x - 2y + 3z = 9
y + 3z = 5
2x - 5y + 5z = 17

inconsistent system
x = -1, y = 1 and z = 2
x = 1, y = -1 and z = -2
x = 1, y = -1 and z = 2

Q 6: Solve the following linear equations.
a + b + c = 10
a + 2b + 3c = 5
2a + b + 5c = 4

a = 27/2, b = -10/2 and c = -7/2
a = 15, b = 2 and c = -8
a = 54/5, b = 17/5 and c = -21/5
inconsistent system

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Question 8: This question is available to subscribers only!


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