Precalculus, Chapter 7, 7.3, Section 7.3, Problem 27

You may use the substitution method to solve the system, hence, you need to use the first equation to write x in terms of z, such that:
2x + 2z = 2 => x + z = 1 => x = 1 - z
You may now replace 1 - z for x in equation 5x + 3y = 4, such that:
5(1 - z) + 3y = 4 =>5 - 5z + 3y = 4 => - 5z + 3y = -1
You may use the third equation, 3y - 4z = 4 , along with -5z + 3y = -1 equation, such that:
3y = 4 + 4z
Replace 4 + 4z for 3y in equation -5z + 3y = -1, such that:
-5z + 4 + 4z = -1 => -z = -1 - 4 => -z = -5 => z = 5
You may replace 5 for z in equation 3y = 4 + 4z:
3y = 4 + 4*5 => 3y = 24 => y = 8
You may replace 5 for z in equation x = 1 - z:

x = 1 - 5 => x = -4
Hence, evaluating the solution to the given system, yields x = -4, y = 8, z = 5.