Is every finitely-presentable group a finite colimit of copies of $F_2$? Let $F_n$ be the free group on $n$ generators. Of course, every finitely-presentable group $G$ is a finite colimit of copies of $F_n$, where $n$ is allowed to vary. But is $G$ a finite colimit of copies of $F_2$?
Of course, because $F_{2n}$ is a finite coproduct of copies of $F_2$, we have that any finitely-presentable group $G$ is a finite colimit of finite colimits of copies of $F_2$ -- a "2-fold" finite colimit of copies of $F_2$. But I'm curious about the 1-fold case. 
To make the question a bit more concrete, let's unwind what it means to be a finite colimit of copies of $F_2$:

Let $G$ be a group. Then $G$ is a finite colimit of copies of copies of $F_2$ if and only if $G$ admits a presentation of the following description:
  
  
*
  
*There are $2n$ generators coming in pairs $x_1,y_1, \dots x_n, y_n$;
  
*There is a finite set of generating relations, each of the form $w(x_i,y_i)=v(x_j,y_j)$, where $w,v$ are group words and $1 \leq i \leq j \leq n$.

So for example, $x_1y_1^2x_1^{-1} = y_2^{-1}x_2$ is a permissible generating relation (with $i=1,j=2$) but $x_1 x_2 = x_3$ is not a permissible generating relation because only 2 different subscripts are allowed to appear in a permissible generating relation. So my question is:
Question: Let $G$ be a finitely-presented group.


*

*Is $G$ a finite colimit of copies of $F_2$?

*Equivalently, does $G$ admit a presentation of the above form?
Edit:
The form of the presentation can be constrained even further, to look like this:


*

*There is a finite set of generating relations, coming in pairs each of the form $x_i = w(x_j,y_j)$, $y_i=v(x_j,y_j)$, where $w,v$ are group words and $1 \leq i, j \leq n$.


Other variations are possible too; I'm not sure what the most convenient description to work with might be.
 A: Here's a pretty direct way to do this. Choose any presentation by generators $x_1,\ldots,x_n$ and relations $r_1,\ldots,r_m$; say $r_i = z_{i,1} \cdots z_{i,k}$ for $z_{i,1},\ldots,z_{i,k} \in \{x_1^{\pm 1}, \ldots, x_n^{\pm 1}\}$. Firstly, we may assume all $r_i$ have length $k = 3$: the new variables
$$x_{i,j} = z_{i,1} \cdots z_{i,j}$$
for $0 \leq j \leq k$ are subject only to the relations
\begin{align*}
x_{i,0} = e = x_{i,k}, & & & & & & x_{i,j} = x_{i,j-1}z_{i,j} & & (1 \leq j \leq k).
\end{align*}
If $k < 3$, we can eliminate the variable $z_{i,1}$ at the expense of replacing all $z_{i,1}^{\pm 1}$ by $z_{i,2}^{\mp 1}$ (if $k = 2$) or $e$ (if $k = 1$) in the other relations, so we may assume all $r_i$ have length exactly $3$. Then introduce new variables $x_{n+1},\ldots,x_{n+m}$ as well as variables $y_1,\ldots,y_{n+m}$, subject to the relations
\begin{align*}
x_{n+i} &= z_{i,1},& & 1 \leq i \leq m,\\
y_i &= e & & 1 \leq i \leq n,\\
y_{n+i} &= z_{i,2}, & & 1 \leq i \leq m,\\
x_{n+i}y_{n+i} &= z_{i,3}^{-1}, & & 1 \leq i \leq m,\\
\end{align*}
This gives a presentation of the desired form. $\square$
A: I believe the answer is yes.  Assume, by way of contradiction, that some finitely presented group cannot be so expressed.  Then we can choose such a group $G$ where for any generating set of the form $x_1,y_1, x_2,y_2,\ldots, x_n,y_n$ the number of non-permissible relations needed to define $G$ (together with some finite number of permissible relations) is minimized; say those non-permissible relations are $w_1=1, w_2=1,\ldots, w_m=1$.  Write $w_1=z_1z_2\cdots z_p$ where each $z_j\in \{x_1^{\pm 1},y_1^{\pm 1},\ldots, x_n^{\pm 1},y_n^{\pm 1}\}$, where we may also assume that $p$ has been minimized.  Note that since $w_1=1$ is not permissible, we must have $p\geq 3$.
Add new generators $x_{n+1},y_{n+1},x_{n+2},y_{n+2}$.  The relations $x_{n+1}=z_{p}$, $y_{n+1}=z_{p-1}$, $x_{n+2}y_{n+2}=1$ are permissible.  The relation $x_{n+2}y_{n+1}x_{n+1}=1$ is also permissible (since it is equivalent to $x_{n+2}=x_{n+1}^{-1}y_{n+1}^{-1}$).  Asserting these relations still gives us the same group (since our new relations merely tell us how to write the new generators in terms of the old ones). The relation $w_1=1$ is equivalent to $z_1z_2\cdots z_{p-2}y_{n+2}=1$, but it is now shorter, a contradiction.
A: Ok, I think the above answers have pointed the way to a proof of the obvious generalization:
Theorem: Consider a variety $V$ (in the sense of universal algebra) generated by operations of arity bounded by some $N \in \mathbb N$, and let $\mathcal C$ be the category of $V$-algebras and homomorphisms. Let $F$ be the free algebra on $N$ generators. Then every finitely-presented $A \in \mathcal C$ is a finite colimit of copies of $F$.
Proof: By using dummy variables, we may assume that every basic operation in $V$ is of arity exactly $N$. As described in the case of groups in the Question, we are looking for a presentation of $A$ by generators $(x_{11}, \dots, x_{1N},\dots, x_{n1}, \dots, x_{nN})$ modulo "permissible" relations $w(x_{i 1}, \dots x_{i N}) = v(x_{j 1},\dots, x_{j N})$, were $w, v$ are (possibly composite) operations in the variety $V$.
As in Pace Nielsen's answer, it will suffice to show that if $w(x_{11},\dots, x_{nN}) = v(x_{11},\dots,x_{nN})$ is a non-permissible relation, we can, after adding more variables $(x_{n+1,1},\dots, x_{n+1,N})$, replace it with relations using only shorter words (in the sense that the total number of basic operations of which each word is composed is smaller -- the base case is a word $x_{ij}$ composed of no operations; note that a relation $x_{ij} = x_{kl}$ is permissible) and a relation of the form $f(x_{n+1,1},\dots, x_{n+1,N}) = v(x_{11},\dots, x_{nN})$.
To this end, we may write $w(x_{11},\dots, x_{nN}) = f(w_1(x_{11},\dots,x_{nN}), \dots, w_N(x_{11},\dots,x_{nN}))$ where $f$ is a basic operation and the $w_i$'s are shorter words. We impose the relations $x_{n+1,i} = w_i(x_{11},\dots,x_{nN})$ (which use only shorter words) along with the relation $f(x_{n+1,1},\dots, x_{n+1,N}) = v(x_{11},\dots, x_{nN})$, as desired.
