In this case, we have the current source $I_s$ entering the junction and the currents $i_1$ and $i_2$ leaving the junction. Therefore, we can write the equation as:
\[I_s = i_1 + i_2\]
Substituting the given value of $I_s = 3A$, we get:
\[3 = i_1 + i_2\]
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