Tables and chairs

Let, the chair and table number respectively $x$ and $y$ .

Then we find two equation,

$2y+1=x\\3(y-1)=x$

Solving that, $x=9$ and $y=4$ .

Adding the chair and table number we find $9+4=\boxed{13}$

let $c$ = number of chairs

let $t$ = number of tables

We have,

$\dfrac{c}{t}=2+\dfrac{1}{t}$ $\Longleftrightarrow$ $\color{#D61F06}\boxed{1}$

$\dfrac{c}{t-1}=3$ $\Longleftrightarrow$ $\color{#D61F06}\boxed{2}$

From $\color{#D61F06}\boxed{2}$ solved for $c$ in terms of $t$ then substitute in $\color{#D61F06}\boxed{1}$ , we have

$\dfrac{c}{t-1}=3$ $\implies$ $c=3t-3$

Then substitute,

$\dfrac{3t-3}{t}=2+\dfrac{1}{t}$

Multiplying both sides by $t$ , we get

$3t-3=2t+1$

$t=4$

It follows that

$c=3(4)-3=9$

The desired answer is $4+9=\color{#3D99F6}\boxed{13}$