1/x-2/y+4=0;1/y-1/z+1=0;2/z+3/x=14 solve by linear equations in three variables
Answers
Answer:
x = 1/2, y = 1/3, z = 1/4
Step-by-step explanation:
Given equations;
1/x-2/y+4=0
1/y-1/z+1=0
2/z+3/x=14
Let;
1/x = a
1/y = b
1/z = c
So, the equations are;
a - 2b + 4 = 0 ........ Eq (1)
b - c + 1 = 0 ........ Eq (2)
2c + 3a - 14 = 0 ........ Eq (3)
Solving Eq (1) & Eq (2):
a - 2b + 4 = 0
b - c + 1 = 0 x 2
a - 2b + 4 = 0
2b - 2c + 2 = 0
----------------------
a - 2c + 6 = 0 .......... Eq (4)
Solving Eq (3) Eq (4):
2c + 3a - 14 = 0
a - 2c + 6 = 0
----------------------
4a - 8 = 0
=> 4a = 8
=> a = 2
Solving Eq (2) & Eq (3):
b - c + 1 = 0 x 2
2c + 3a - 14 = 0
2b - 2c + 2 = 0
2c + 3a - 14 = 0
-----------------------
3a + 2b - 12 = 0
2b = 12 - 3a
Substituting the value of 'a' in the above equation;
2b = 12 - 3(2)
=> 2b = 12 - 6
=> 2b = 6
=> b = 3
From Eq (2);
b - c + 1 = 0
Substituting the value of 'b' in the above equation;
3 - c + 1 = 0
=> c = 4
1/x = a 1/y = b 1/z = c
1/x = 2 1/y = 3 1/z = 4
Therefore, the required solution is;
x = 1/2, y = 1/3, z = 1/4
Answer:
I have a doute first equation 1+2 next 2+3 and next equation I not understand please write and send the same sum please