I'll give brainliest award to anyone who shows good ! explain how to solve the following system of equation. what is the solution to the system? 2x+2y+z=-5 3x+4y+2z=0 x+3y+2z=1

Eliminate z from equation 2 and 3: subtract ③ from ②: 2x+y=-1 and call this ④.
Double ①: 4x+4y+2z=-10 and call it ⑤.
⑤-②: x=-10.
Substitute x=-10 in ④: -20+y=-1, y=19.
Substitute for x and y in ①: -20+38+z=-5, z=-23.
Solution: x=-10, y=19, z=-23.

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.