Answer:
x = (12-k)/2, y = k, z = (k-6)/4
Explanation:
Given the system of equation
x + y - 2z = 9 ... 1
3x + y + 2z = 15 ...2
x - 5y + 22z = -27... 3
First let us reduce the system of equation into two with two unknowns.
Subtracting 1 from 3
y-(-5y) + (-2z-22z) = 9-(-27)
y+5y + (-24z) = 9+27
6y-24z = 36 ... 4
Multiplying equation 1 by 3 and subtracting from equation 2
3x + 3y - 6z = 27
3x + y + 2z = 15
On subtracting both;
(3y-y)+(-6z-2z) = 27-15
2y-8z = 12 ... 5
Equating 4 and 5
6y-24z = 36 ... 4
2y-8z = 12 ... 5
Multiplying equation 5 by 3 the equation becomes;
6y-24z = 36 ... 6
6y-24z = 36 ... 7
We can see that equation 6 and 7 are the same;
let y = k
6k - 24z = 36
k - 4z = 6
4z = k-6
z = k-6/4
Substituting y = k and z = k-6/4 into equation 1 to get x
From 1; x + y - 2z = 9 ... 1
x + k -2( k-6/4) = 9
x + k - (k-6)/2 = 9
x = 9+(k-6)/2-k
x = {18+(k-6)-2k}/2
x = (12-k)/2
The solutions to the system of equations are x = (12-k)/2, y = k, z = (k-6)/4 where k is any constant. This shows that the system of equation has infinite solutions.