what is the solution to the system of equations -3x-y+z=6 -3x-y+3z=0 x-3z=0

Question

3x-y+z=6

-3x-y+3z=0

x -3z=3

Answer 1

The equations in the title differ from those in the context. Starting with the equations in the title and subtracting equation 1 from 2 we get 2z=-6, so z=-3. Eqn 3 gives x=3z=-9. Eqn 1 gives y=-3x+z-6=27-3-6=18. Check for consistency by substituting these values in each equation. All equations are OK.

Now the context equations.

Add eqns 1 and 2: -2y+4z=6 or -y+2z=3. From this z=(3+y)/2.

Add eqns 2 and 3: -2x-y=3 or 2x+y=-3. From this x=-(3+y)/2=-z.

From eqn 3: x-3z=4x=3, so x=3/4. Therefore, z=-3/4 and y=-3-2x=-3-3/2=-9/2..

Solution: x=3/4, y=-9/2, z=-3/4. These fit with the equations in the context.