solving the system of linear equation; samlpe solution x=2, y=3, z=0

Question

how do we solving the x+2y-z=3, -3x+y-4z=-1, -3x+2y-2=-2 can you show me any method to make me understand

Answer 1

x+2y-z=3, -3x+y-4z=-1, -3x+2y-2=-2

First of all, is the last equation: -3x + 2y -z = -2 or is it -3x + 2y - 2 = -2?

for this, I would use the system of linear equations and solve it by first eliminating whatever I can using one equation with another; then substitute where I can.

1)   -3x + y - 4z = -1  <------ multiply this by - 1

      -3x + 2y - z = -2  (if the last term is z)

     -----------------------

       3x -   y + 4z = 1

-3x + 2y -   z = -2

----------------------

 

Answer 2

the sample solution is not a solution.

 

x+2y-z=3  using the sample solution

 

2+6-0=?3

 

8≠3

Thus the sample solution is not a solution.

A set of three independent equations can be solved for three variables.   In this case the solution involves fractions, which makes me suspect that the problem is not written correctly.

There is no point in solving this problem until one confirms that it is written correctly.