Mesh current method in finding currents

The equation for clockwise should be:

$$-v_{s} + (i_1 + i_s)R_2 + i_1R_3 = 0$$ $$-v_s + i_1(R_2+R_3) + i_sR_2 = 0$$ $$i_1 = - \frac{ i_sR_2 - v_s}{R_2+R_3}$$

Which is exactly negative of the other direction choice.

None of the given answers really address the question. The mesh analysis approach allows us to reduce the number of degrees of freedom significantly. In the example, mesh analysis provides 3 variables while there are 10 total edges in the circuit. So clearly, we could get and solve 10 equations and 10 variables, but mesh analysis tells us we only need to solve 3.

Two facts are true to allow this reduction.

1. Using the mesh currents approach will satisfy the node and loop rules
2. A set of currents and voltages that obey both rules is unique.

The first is easy to see, but the second explains why mesh analysis works. Since the mesh currents satisfy the node and loop rules, it must be the only solution.

It is not straightforward to see why statement 2 must be true. The proof by contradiction assumes that two such current-voltage arrangements exist satisfying the node and loop rules. It then shows that you can find a loop where the sum of voltage drops is non-zero. https://math.stackexchange.com/questions/1742680/unique-solution-for-circuits-in-linear-algebra