In November 2003, the Four Companies with the Largest Number$\begin{array} { l }
z - u = 3 ( x - y ) + 6 \
x + y = 52 + z + u \
x - y + z - u = 42
\end{array}$

Question

In November 2003, the Four Companies with the Largest Number$\begin{array} { l } 
z - u = 3 ( x - y ) + 6 \
x + y = 52 + z + u \
x - y + z - u = 42
\end{array}$

Examlex · Accepted Answer

The Answer of In November 2003, the Four Companies with the Largest Number$\begin{array} { l } 
z - u = 3 ( x - y ) + 6 \
x + y = 52 + z + u \
x - y + z - u = 42
\end{array}$

In November 2003, the Four Companies with the Largest Number $\begin{array} { l } z - u = 3 ( x - y ) + 6 \\x + y = 52 + z + u \\x - y + z - u = 42\end{array}$

Definitions:

In November 2003, the Four Companies with the Largest Number z−u=3(x−y)+6x+y=52+z+ux−y+z−u=42\begin{array} { l } z - u = 3 ( x - y ) + 6 \\x + y = 52 + z + u \\x - y + z - u = 42\end{array}z−u=3(x−y)+6x+y=52+z+ux−y+z−u=42​

Definitions:

In November 2003, the Four Companies with the Largest Number $\begin{array} { l } z - u = 3 ( x - y ) + 6 \\x + y = 52 + z + u \\x - y + z - u = 42\end{array}$