Solve the system. If there is more than one solution, write the general solution. x + y - 2z = 9 3x + y + 2z = 15 x - 5y + 22z = -27 Solution {-2z + 3, 4z + 6, z | z is any real number}

Step-by-step 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.

step-by-step explanation: