Solve the system of linear equations and check any solutions algebraically. (If there is no solution, enter NO SOLUTION. If the system is dependent, express x, y, and z in terms of the parameter a.)

x +2z=5
3x-y-z=12
6x-y-5z=27

in Matrix form:

1,0,2 P5
3,-1,-1:12
6,-1,-5:27

so subtracting equation 2 from 3
we get equation 1, and
3,0,-4:15
Now multiply equation 1 by a factor of 2 to get
2,0,4:10
now adding those two equations
5,0,0:25 orx=5
putting that into equation 1, we get z=0
putting those into equation 2 we get
15-y-0=12 or y=3
and that checks in equation 3.

subtract the 2nd equation from the 3rd equation

solve the two remaining x and z equations with substitution or elimination

x + 2z = 5

3x - 4z = 15

but what are the parameters of a?

only if the system is dependent ... which it isn't

To solve the system of linear equations, we will use the method of elimination or substitution. Let's start with the elimination method.

Step 1: Multiply the first equation by 3 and the second equation by 1 to make the coefficients of x in both equations opposite and equal.
3(x + 2z) = 3(5) => 3x + 6z = 15
1(3x - y - z) = 1(12) => 3x - y - z = 12

Step 2: Now, subtract the second equation from the first equation to eliminate x.
(3x + 6z) - (3x - y - z) = 15 - 12
3x - 3x + 6z + y + z = 3
7z + y = 3 => y = -7z + 3 (Equation 3)

Step 3: Multiply the second equation by 6 and the third equation by 1 to make the coefficients of x in both equations opposite and equal.
6(3x - y - z) = 6(12) => 18x - 6y - 6z = 72
1(6x - y - 5z) = 1(27) => 6x - y - 5z = 27

Step 4: Now, subtract the third equation from the second equation to eliminate x.
(18x - 6y - 6z) - (6x - y - 5z) = 72 - 27
18x - 6x - 6y + y - 6z + 5z = 45
12x - 5y - z = 45 => 12x = 5y + z + 45 (Equation 4)

Step 5: Substitute y = -7z + 3 (from Equation 3) into Equation 4.
12x = 5(-7z + 3) + z + 45
12x = -35z + 15 + z + 45
12x = -34z + 60
x = (-34z + 60)/12
x = -17/6z + 5 (Equation 5)

Therefore, the solution to the system of linear equations is:
x = -17/6z + 5
y = -7z + 3
z = z

To check the solutions algebraically, substitute the values of x, y, and z into the original equations and see if the equations hold true:

For the first equation:
x + 2z = 5
((-17/6z) + 5) + 2z = 5
(-17z + 30 + 12z) / 6 = 5
-5z + 30 = 30
-5z = 0
z = 0

For the second equation:
3x - y - z = 12
3((-17/6z) + 5) - (-7z + 3) - z = 12
((-17/2z) + 15) + 7z - 3 - z = 12
(-17 + 30z + 14 - 6z) / 2 = 12
8z - 3 = 24
8z = 27
z = 3.375

For the third equation:
6x - y - 5z = 27
6((-17/6z) + 5) - (-7z + 3) - 5z = 27
((-17z + 30) - (-7z + 3) - 5z) = 27
(-17z + 30 + 7z - 3 - 5z) = 27
-15z + 27 = 27
-15z = 0
z = 0

Therefore, the solution to the system of linear equations is:
x = -17/6(0) + 5 => x = 5
y = -7(0) + 3 => y = 3
z = 0

Hence, the solution to the system of linear equations is x = 5, y = 3, and z = 0.