\[\begin{align}
x+x+y&=14\\
L+L+z&=20\\
x+y+z&=12\\
x+z+L&=18
\end{align}\]
If numbers $x,y,z,L$ satisfy the equations above, find the value of $x+z$.
+267 after bashing randomly, I got x=6, y=2, z=4, and L=8.
since x=6 and z=4, 6+4=10
+128539 2x + y = 14 ⇒ y = 14 - 2x
2L + z = 20 ⇒ z = 20 - 2L
x + y + z = 12
x + z + L = 18
Using the last two equations and substituting
x + 14 - 2x + 20 - 2L = 12
-x - 2L = -22 (1)
x + (20- 2L) + L =18
x - L = -2 (2)
Add (1) and (2)
-3L = -24
L = 8
x - 8 = -2
x = 6
z = 20 - 2L = 4
y = 14 - 2x = 2
x + z = 10
