3x + 5y = 1 + z,        (1)
4x + 7y = 2 - z.        (2)


I will solve the system (1), (2) for x using Elimination. 
For it, multiply eq.(1) by 4 (both sides). Multiply eq.(2) by 3 (both sides). You will get an equivalent system

12x + 20y = 4 + 4z,     (3)
12x + 21y = 6 - 3z      (4)

Subtract (3) from (4). You will get

 y = -7z + 2            (5)


Now, I will solve the system (1), (2) for y using Elimination. 
For it, multiply eq.(1) by 7 (both sides). Multiply eq.(2) by 5 (both sides). You will get an equivalent system

21x + 35y =  7 + 7z,    (6)
20x + 35y = 10 - 5z     (7)

Subtract (7) from (6). You will get

 x = 12z - 3            (8)


The given system of two equations in three unknowns has infinitely many solutions.
You can take z as an arbitrary real number and calculate x and y according to the formulas (5) and (8):

 x =  12z - 3,
 y =  -7z + 2.