Cheating in (2,6)-Shamir Sharing Scheme
You have four equations and two unknowns.  Any two of those equations should therefore be solvable, and all pairs that don't include the cheater should yield the same solution.

(You don't actually need to solve all six possible pairs of equations if you know there's at most one cheater.  It's sufficient to e.g. just solve equations 1 &amp; 2, and check which of 3 &amp; 4 fails to fit the solution.  If both fail, solve 3 &amp; 4 instead, and check which of 1 &amp; 2 fails to fit.)

Also, remember that you will need to solve all those equations in $\mathbb Z_{11}$, i.e. using arithmetic modulo 11.