Since the third equation contains x an y but not z, eliminate z from the first two equations by adding twice the first equation to 5 times the second equation. The result is 31x + 14y = 76.
Now solve 31x + 14y = 76 and 7x + 5y = 18 by 5 times the first equation to -14 times the second equation. The result is 57x = 128. Then x = 128/57.
Substitute this into 7x + 5y = 18 to find y. Then substitute for both x and y in 3x - 3y + 5z = 13 to find z.
