What is the solution to the system of equations?2x−y+z=−8 x+y+z=−4 3x−y−z=−4

The solution to this system is (-2, 1, -3).

Step-by-step explanation:

Let's eliminate variable x first.  Combine the first two equations, obtaining:

3x - 0y + 2z  = -12.

Now subtract the third equation from this result:

 3x - 0y + 2z  = -12

-(3x −   y −   z =  −4)

           y + 3z  = -8

Similarly, combine the second and third original equations to eliminate x again.  To do this, subtract  2(x + y + z = -4) from the first equation:

2x−y+z=−8

-2x - 2y - 2z = 8

    -3y - z = 0

Now we have eliminated x completely, and find from -3y - z = 0 that z = -3y.  Substitute this -3y for z in the equation y + 3z  = -8 found above:

y + 3(-3y)  = -8.  Then y - 9y = -8, and so y must = 1.  From -3y - z = 0, substituting 1 for y, we find that z = -3(1), or z = -3.

Finally, subst. 1 for y and -3 for z in the second equation:

x + 1 - 3 = -4

So, x - 2 = -4, and thus x must be -2.

The solution to this system is (-2, 1, -3).

idk

step-by-step explanation: