Given:

2x-3y+z=0

3x+2y=35

4y-2z=14

Which of the following is a solution to the given system?

(2, 3, 5)

(3, 2, 0)

(1, 16, 0)

(7, 7, 7)

Respuesta :

If x and y are either 2 or 3 it would not fit 3x+2y=35.
so the first two solutions are out.
Try the third one 16*4 is 64. 64-0 is not 14 so the third one is out too.

The answer is (7,7,7)

Answer:

the answer is (7, 7, 7)

Step-by-step explanation: