Precalculus, Chapter 7, 7.3, Section 7.3, Problem 44

You should notice that the system is indeterminate, since the number of variables is larger than the number of equations.
2*(2x + 3y + 3z) = 14 => 4x + 6y + 6z = 14 => 4x = 14 - 6y - 6z
Replace 14 - 6y - 6z for 4x in the second equation, such that:
4x = 44 - 18y - 15z => 14 - 6y - 6z = 44 - 18y - 15z
12y + 9z = 30 => 4y + 3z = 10 => 4y = 10 - 3z => y = 5/2 - (3/4)z
Replace back 5/2 - (3/4)z for y in equation 4x = 14 - 6y - 6z , such that:
4x = 14 - 6( 5/2 - (3/4)z ) - 6z
4x = 14 - 15 + (9/2)z - 6z
4x = -1 - (3/2)z => x = -1/4 - (3/8)z
Hence, evaluating the solutions to the system, yields x = -1/4 - (3/8)z, y = 5/2 - (3/4)z, z = z.