= Solution
For an <algebra for a monad> $(A,a)$, consider the fork in the <Eilenberg-Moore category>
$$
\boxed{F(TA)\ \mathrel{\substack{\xrightarrow{\ \mu_A\ }\\[-2pt]\xrightarrow[\ Ta\ ]{}}}\ F(A)\xrightarrow{\ a\ }(A,a).}
$$
Here $\mu_A:F(TA)\to F(A)$ is the counit at $F(A)$, and $Ta=F(a)$. The arrow $a$ is an algebra morphism by $a\mu_A=aTa$, which also says that it coequalizes the two arrows.
Let $h:F(A)\to(B,b)$ be an algebra morphism with $h\mu_A=hTa$. Define $g=h\eta_A:A\to B$. Naturality of $\eta$ at $a$ gives
$$
ga=h\eta_Aa=hTa\,\eta_{TA}=h\mu_A\eta_{TA}=h.
$$
Since $h$ is an algebra morphism,
$$
bTg=bTh\,T\eta_A=h\mu_A T\eta_A=h=ga.
$$
Thus $g:(A,a)\to(B,b)$ is an algebra morphism with $ga=h$. Any other such factorization satisfies $g'=g'a\eta_A=h\eta_A=g$. This proves the full <coequalizer> universal property inside the algebra category.
The pair is moreover a <reflexive pair>: its common section is $F(\eta_A)$, with underlying map $T\eta_A$, because $\mu_AT\eta_A=1_{TA}$ and $Ta\,T\eta_A=T(a\eta_A)=1_{TA}$. Hence \b[every monad algebra is a reflexive coequalizer of free algebras]. This <reflexive free-algebra presentation of a monad algebra> needs no general existence theorem for arbitrary algebra-category colimits.
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