What is the correct number of solutions for the following system of equations?

2x + y - 3z = 4

 4x + 2z = 10

-2x + 3y - 13z = -8

y = 3z - 2x + 4

3y = 9z - 6x + 12

-2x + 9z - 6x + 12 - 13z = -8

-8x - 4z = -20

8x + 4z = 20

2x + z = 5 

If z = 1, x = 2

In that case y = 3

Answer choice:

C. One Solution

4x + 2z = 10 --> 4x = 10 - 2z --> x = (5 - z)/2.  Or 4x + 2z = 10 --> 2z = 10 - 4x --> z = 5 - 2x.

If you substitute for z in equation 1 or 3, you got 8x + y = 19 and subsequently y = 19 - 8x.  Substituting for x in those equations gets you y - 4z = -1 and subsequently y = 4z - 1.

az is right: there are infinitely many solutions.  A).

Eqn 2 implies 2x = 5 - z.

Then Eqn 1 says

5 - z + y - 3z = 4, or y - 4z = -1.

And Eqn 3 says

z - 5 + 3y - 13z = -8, or 3y - 12z = -3, or

y - 4z = -1.

So you could choose ANY value for z, and

y would be 4z - 1, and x would be 5 - z.

Answer:  A, Infinitely many solutions.

  2x + y - 3z = 4

   y = 3z - 2x + 4  

  4x + 2z = 10 

  -2x + 3y - 13z = -8 

  -2x + 3(3z - 2x + 4) - 13z = -8 

 8x + 4z = 20 

 4x + 2z = 10 

 2x + z = 5 

  x = 2, z = 1 y = 3 

 Answer choice:  

 C. One Solution