Question 1 :
Solve the following system of linear equations in three variables
x + y + z = 5 ; 2x − y + z = 9 ; x − 2y + 3z = 16
Solution :
x + y + z = 5 -------(1)
2x − y + z = 9 -------(2)
x − 2y + 3z = 16 -------(3)
Let us add (1) and (2) |
Multiply the first equation by 2 and add by (3) |
Let 3x + 2z = 14 ----(4) and 3x + 5z = 26 ---(5)
(4) - (5)
3x + 2z = 14
3x + 5z = 26
(-) (-) (-)
----------------
-3z = -12 ==> z = 4
By applying z = 4 in (4), we get
3x + 2(4) = 14
3x + 8 = 12
3x = 14 - 8
3x = 6 ==> x = 2
By applying x 2 and z = 4 in (1), we get
2 + y + 4 = 5
6 + y = 5
y = 5 - 6
y = -1
Hence the solution is
(x, y, z) = (2, -1, 2)
Question 2 :
(1/x) - (2/y) + 4 = 0
(1/y) - (1/z) + 1 = 0
(2/z) + (3/x) = 14
Solution :
Let 1/x = a, 1/y = b and 1/z = c
a - 2b = -4 ----(1)
b - c = -1 ----(2)
3a + 2c = 14 ----(3)
(1) + 2(2) ==> a - 2b + 0c = -4
0a + 2b - 2c = -2
---------------------
a - 2c = -6 ----(4)
(4) + (3)
a - 2c = -6
3a + 2c = 14
----------------
4a = 8 ==> a = 2
By applying a = 2 in (4), we get
2 - 2c = -6
-2c = -6 - 2
-2c = -8 ==> c = 4
By applying the value of a in (1),
a - 2b = -4 ----(1)
2 - 2b = -4
-2b = -4 -2
-2b = -6 ==> b = 3
a = 1/x If a = 2 1/x = 2 x = 1/2 |
b = 1/y If b = 3 1/y = 3 y = 1/3 |
b = 1/z If c = 4 1/z = 4 z = 1/4 |
Hence the solution is
(x, y, z) = (1/2, 1/3, 1/4)
Question 3 :
x + 20 = (3y/2) + 10 = 2z + 5 = 110 - (y + z)
Solution :
x + 20 = (3y/2) + 10 x + 20 = (3y + 20)/2 2x + 40 = 3y + 20 2x - 3y = 20 - 40 2x - 3y = -20 -----(1) |
(3y/2) + 10 = 2z + 5 3y + 20 = 4z + 10 3y - 4z = 10 - 20 3y - 4z = -10 -----(2) |
2z + 5 = 110 - (y + z)
2z + 5 = 110 - y - z
2z + z + y = 110 - 5
3z + y = 105 ----(3)
(1) + (2)
2x - 3y + 0z = -20
0x + 3y - 4z = -10
------------------------
2x - 4z = -30 ---(4)
(1) ==> 2x - 3y + 0z = -20
3(3)==> 0x + 3y + 9z = 315
------------------------
2x + 9z = 295 ----(5)
(4) - (5)
2x - 4z = -30
-2x - 9z = -295
-------------------
-13z = -325
z = 25
By applying the value of z in (4), we get
2x - 4(25) = -30
2x - 100 = -30
2x = -30 + 100 = 70
x = 35
By applying the value of z in (3), we get
3(25) + y = 105
75 + y = 105
y = 105 - 75
y = 30
Hence the solution is
(x, y, z) = (35, 30, 25)
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