Nodal analysis of operation amplifier
The op-amp symbol can sometimes lead one astray as, I believe, it has in this case.

Recall that the output of an ideal op-amp is an ideal voltage controlled voltage source which means that the output will source or sink any amount of current.

Let's redraw the circuit by replacing the op-amp with an ideal amplifier model:







simulate this circuit &ndash; Schematic created using CircuitLab

By voltage division and superposition, we write

$$V_- = V_i \frac{R_2}{R_1 + R_2} + V_o\frac{R_1}{R_1 + R_2}$$

But, by inspection

$$V_o = A(V_+ - V_-) = -AV_-$$

thus

$$V_o = - \frac{R_2}{R_1 + \frac{R_1 + R_2}{A}}V_i$$

For an ideal op-amp, \$A \rightarrow \infty\$ so

$$V_o = -\frac{R_2}{R_1}V_i $$

Now, this is not the typical method to solve op-amp problems.  However, I hope this gives you some insight into why your initial equations are incorrect.