I don't know of a site that explains how to do a super mesh but I can walk you through it . (if anyone sees a mistake please tell me as it's been a while since I've done that)First of all in the diagram you've given us it's clear that Ia = -1A because it's the only current through the current source on the left and it's direction is opposite the the one drawn for Ia. We can also state that :Ib -Ic = 1A because they both go through the current source to the right in opposite directions (it's actually simpler to understand if we write : Ib + (-Ic) = 1A)Now, we need one more equation to solve for Ib and Ic.We need to create a voltage loop that we walk around a full round.I think the simplest one here is the one that includes the two right most parts of the circuit :And that will be
We go clockwise starting from above the right most resistor) :-1*Ic -2 -1*Ib -1 -1*Ib = 0 When we pass a resistor in the direction of the current we go down in the voltage (so we add it as a negative term that equals the total current that goes through it times the resistance). When we pass a voltage source from the (+) to the (-) we also subtract it's size and vice verse. Now if we take the three constraints we have and solve them we'll get all three currents . and of course :Ix = Ia + (-Ib) .Good Luck!