System

Solve the following system:
{y + 2 x = 14 | (equation 1)
z + 2 L = 20 | (equation 2)
x + y + z = 14 | (equation 3)
L + x + z = 18 | (equation 4)

Subtract 1/2 × (equation 1) from equation 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y/2 + z + 0 L = 7 | (equation 3)
x + 0 y + z + L = 18 | (equation 4)

Multiply equation 3 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y + 2 z + 0 L = 14 | (equation 3)
x + 0 y + z + L = 18 | (equation 4)

Subtract 1/2 × (equation 1) from equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y + 2 z + 0 L = 14 | (equation 3)
0 x - y/2 + z + L = 11 | (equation 4)

Multiply equation 4 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y + 2 z + 0 L = 14 | (equation 3)
0 x - y + 2 z + 2 L = 22 | (equation 4)

Swap equation 2 with equation 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 2 L = 20 | (equation 3)
0 x - y + 2 z + 2 L = 22 | (equation 4)

Add equation 2 to equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 2 L = 20 | (equation 3)
0 x + 0 y + 4 z + 2 L = 36 | (equation 4)

Divide equation 4 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 2 L = 20 | (equation 3)
0 x + 0 y + 2 z + L = 18 | (equation 4)

Swap equation 3 with equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + z + 2 L = 20 | (equation 4)

Subtract 1/2 × (equation 3) from equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + 0 z + (3 L)/2 = 11 | (equation 4)

Multiply equation 4 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + 0 z + 3 L = 22 | (equation 4)

Divide equation 4 by 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Subtract equation 4 from equation 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + 0 L = 32/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Divide equation 3 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Subtract 2 × (equation 3) from equation 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 0 z + 0 L = 10/3 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Subtract equation 2 from equation 1:
{2 x + 0 y + 0 z + 0 L = 32/3 | (equation 1)
0 x + y + 0 z + 0 L = 10/3 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Divide equation 1 by 2:
{x + 0 y + 0 z + 0 L = 16/3 | (equation 1)
0 x + y + 0 z + 0 L = 10/3 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Collect results:

 x = 16/3
y = 10/3
z = 16/3
L = 22/3                            x  +  z ==16/3  +  16/3==32/3

We have the system of equations: 

\(2x + y = 14\)

\(2L + z = 20\)

\(x +y + z = 14\)

\(x + z + L = 18\)

Subtracting \(y\) from both the first and third equations gives us: \(x + z = 14 - y\) and \(2x = 14-y\)

Because both equations are equal to each other, we can set them equal. This gives us: \(x + z = 2x\), meaning \(x = z \)

Substituting this in, we have a new system in terms of \(x\):

 \(2x + y = 14\)

\(2L + x = 20\)

\(2x + L = 18\)

Now, we are looking to find the value of \(2x\). 

We can do this by disregarding the first equation, which gives us this system: 

\(2L + x = 20\)

\(2x + L = 18\)

Multiplying the second equation by 2 gives us: \(4x + 2L = 36\)

Now, subtracting the first equation from this shows: \(3x = 16\).

Multiplying both sides by \(2 \over 3\) shows \(x +z = \color{brown}\boxed{32 \over 3}\), just as Guest found