Hello, dhiab!
These are strange problems . . .
Solve in $\displaystyle N^2\!:\quad\begin{array}{ccc}xy & \leq & 2x \\ x+y &=& 4\end{array}$
The line $\displaystyle x + y \;=\:4$ has intercepts (4,0) and (0,4). Code:

* 
*
 *
 *
 *
  +    *  
 *

We have the inequality: .$\displaystyle xy  2x \:\leq\:0 \quad\Rightarrow\quad x(y2) \:\leq\:0 $
This is true when $\displaystyle x$ and $\displaystyle (y2)$ have opposite signs.
There are two cases: .$\displaystyle \begin{Bmatrix}x \:\geq \:0 \\ y \:\leq \:2\end{Bmatrix}\;\text{ and }\;\begin{Bmatrix}x \:\leq\:0 \\ y \:\geq \:2\end{Bmatrix}$
The graph looks like this: Code:
::::::::
::::::::
::::::::
::::::::
    2+      
::::::::::
+:::::
::::::::::
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Together, we have this graph . . . Code:
::::*:::
::::::*:
::::::::*
:::::::: *
    2+  *    
:::::*:::::::
+::::*:::
:::::::::*:::
:::::::::::*:
:::::::::::::*
The final graph looks like this: Code:
* 
* 
*

2+  *    
 *
+*
 *
 *
 *
How can we describe this graph?
Maybe: .$\displaystyle x + y \:=\:4\:\text{ for }(x \leq 0) \cup (x \geq 2)$