In the picture.
Due to the series configuration, Ic and Id are equal in magnitude but opposite in direction.
if Ia = 3A, Ic = 1A, then Id = -1A
Ia + Ic - Id - Ib = 0
3A + 1A - 1A - Ib = 0
as a result, Ia=Ib = 3A
I agree with the Ic Id sign. However, Using the Node with Ia and Ic entering and Ib leaving and using KCL. I get Ia+Ic-Ib=0 therefore Ia+Ic=Ib or 3A + 1A =4A
You are right.
for the node with Ib and Id entering, Ia leaving
-Ia + Ib + Id = 0
-3 + Ib + -1 = 0
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