Your diagram makes me nuts. I hope I never see another one like this.
If you assume that the wire carrying I8 is a single node, then I4 is in parallel with I3. The voltage sources look like a short to the current sources, so the resistor is shorted out, so now you have I1 + I2 = I3 + I4, so I4 must be zero, so open the line carrying I4.
Notice that I1, I5 and I6 are all going into the same node. This is impossible, so the sign of one of these currents is wrong.
Since I4 = 0, I5 = I8.
Since the voltage of Uq1>the voltage of Uq2, I think I8 is the correct direction of current so the I5 arrow should be reversed. This doesn't seem to contradict anything so far. Delete the I5 notation from the drawing.
So,
I8 = I1 + I6
I8 = I3 + I7
I6 = I2 + I7
I1 = 10A
I2 = 20A
I3 = 30A
I couldn't solve this in closed form so I put it into a spreadsheet and plugged in numbers for I7.
When I plugged in 1A for I7, I8 said 31A, 31A and I6 said 21A.
With I7 at 100A, I8 said 130A, 31A and I6 said 120A.
The I1-I3 totals always worked.
I think I7 is indeterminate without knowing the value of R. It is a circulating current and is due to the voltage sources and the resistor.
I think the Uq1, Uq2, R circuit counts as a single node for the purposes of I1-I3.
Somewhere in all this is the answer to your question, I hope.