Is there a systematic way to solve equations in the braid groups? In particular, if $B_3$ is the braid group on three strands with the presentation $\{ a,b\  \ aba = bab \}$, how do I find $x$ so that $xaxbx^{1}b^{1}x^{1} = bax^{1}b^{1}x$ (if such an $x$ exists)?

This group is a central extension of $PSL_2\mathbb{Z}$, which is a virtually free group. There is an algorithm to solve equations in such groups, and parameterize the solutions. Since your equation is degree zero in $a,b,x$, if the lift of the solution in $PSL_2\mathbb{Z}$ to $B_3$ solves the equation for one lift, it should work for any other lift. I'm not quite sure though how to determine this uniformly over all lifts of the solution. The solutions are given by MakaninRazborov diagrams, and they are parameterized by various automorphisms. So I think you just need to check one solution in each equivalence class coming from each orbit. 


The braid group is automatic, so has a solvable word problem. You might find this helpful. However, it's not clear to me that this means that equations can actually be solved. 


Your equation is $xaxbXBXXbxAB = 1$; thus the substitution $x = b$ answers your particular question. May I ask where this equation comes from? 

