Since you wrote the equations we can verify them together as follows.
If you like writing each equation as sum of V's on the left side and sum of I*R's on the right side, we can follow the following rules:
(1) The voltage sign of a source is positive if the loop current flows from the minus terminal to the positive terminal of its source. It is negative otherwise.
(2) If all loop currents are made of the same direction (clockwise or counterclockwise) and a resistor R is common with an adjacent loop
, the voltage on R would be [ + ( I - I_n ) * R ], where I is the current of the equation's loop. Otherwise it is (+ I * R).
By following the above rules, we get:
For I_1 loop : V3 = I_1 * R12 + ( I_1 - I_3) * R8
For I_2 loop : V1 + V4 = I_2 * R13 + (I_2 - I_3) * R11
For I_3 loop : - V2 - V4 = (I_3 - I_2) * R11 + I_3 * R9 + ( I_3 - I_1 ) * R8
As you noticed, you did a few errors related to sign.
Now we can re-write these 3 equations after substituting the given values.
And we get 3 equations with 3 unknowns. I think your learnt how to solve them.
Note:
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