Question

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

Answer 1

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.

Answer 2

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

Answer 3

but what are the parameters of a?

Answer 4

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