2x+y=14 (1)
L+L+z=20 (2)
x+y+z=12 (3)
x+z+L=18 (4)
\(0L+2x+y+0z=14\)
\(2L+0x+0y+z=20\)
\(0L+x+y+z=12\)
\(L+x+0y+z=18\)
Add the 4 equations
\(3L+4x+2y+3z=64\)
3(L+z)+2(2x+y)=64 I factorising by grouping.
We know from (1) that 2x+y=14
so
3(L+z)+2(14)=64
3(L+z)+28=64 I subtract 28
3(L+z)=36 I divide by 3
L+z=12 (5)
So we can find the value of L from (2)
L+L+z=20
we know that L+z=12 from (5)
L+12=20
L=8
Well also from (2) since we knew that L=8 we can now z
8+8+z=20
z=20-16
z=4
from (4) we know that x+z+L=18
so
x+4+8=18
x=18-12
x=6
The question may end here as you asked "Find x+z" we know x and we know z so x+z=6+4=10 , the answer is 10.
However, let's contuine to get the value of y.
from (1) we know that x+x+y=14
we know that x=6
thus,
6+6+y=14
y=14-12
y=2
So,
L=8
x=6
y=2
z=4
Confirmation from Wolfram:
https://www.wolframalpha.com/input/?i=systems+of+equations+calculator&assumption=%22FSelect%22+-%3E+%7B%7B%22SolveSystemOf4EquationsCalculator%22%7D%7D&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation1%22%7D+-%3E%222x%2By%3D14%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation2%22%7D+-%3E%22L%2BL%2Bz%3D20%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation3%22%7D+-%3E%22x%2By%2Bz%3D12%22&assumption=%7B%22F%22%2C+%22SolveSystemOf4EquationsCalculator%22%2C+%22equation4%22%7D+-%3E%22x%2BL%2Bz%3D18%22