I'm using biconditional graphs for the lack of a better name, I don't know the name of it, if you know the real name, feel free to edit it.
Draw the set of points given by: $|x-3|\leq1$ and $|y-2|\leq5$
I have evaluated the values for which the conditions are true and I obtained $\{x:2\leq x\leq 4\}$ and $\{y:-3\leq y\leq 7 \}$, now I'm trying to plot as if it were a function $f(x):=|x-3|\leq1$, then I'm using the values of $y$ for making a correlation with $x$, for example:
$$\begin{eqnarray*} {f(2)}&:=&{|2-3|\leq1} \\ {}&:=&{1\leq 1 } \end{eqnarray*}\tag{1}$$
$$\begin{eqnarray*} {f(3)}&:=&{|3-3|\leq1} \\ {}&:=&{0\leq 1 } \end{eqnarray*}\tag{2}$$
$$\begin{eqnarray*} {f(4)}&:=&{|4-3|\leq1} \\ {}&:=&{1\leq 1 } \end{eqnarray*}\tag{3}$$
Thus $f(2)=1$, $f(3)=0$, $f(4)=1$, but I can't find a match, because there are not $0$'s nor $1$'s in the interval $2\leq x \leq 4$, is this correct?
The secondary hypothesis I have is that this kind of operation is made to plot planes, not lines, then I should just draw a plane, is that it?
Note that you have two inequalities: solving one gives you the values of $x$ defined by the first inequality; solving the other gives you the values of $y$ defined by the second inequality. Neither inequality is a function, and so $x$ is certainly not a function of $y$, nor is $y$ a function of $x$.
Try graphing the lines $y = -3$ and $y = 7$: those will be the horizontal bounds of the region defined by your inequalities. Then graph the the vertical lines $x = 2, x = 4$: giving you the bounded region of interest: the rectangular region with vertices: $$(2,-3),(2,7),(4,-3),(4,7)$$
Then simply "shade" the region bounded by these lines, inclusive of the lines.
From Wolfram Alpha (though note the scale for the x, y axis are not matched. If you were to graph this to scale, you'd have a rectangle that is much "taller" than it is wide):
ADDED: Nicer graph of the inequalities (in that it is properly scaled):