Weighted summing amplifier

$i_1 = \frac{V_1-V_2}{R_1+R_2}$ $V_+ = V_- = V_1 - i_1R_1 = V_1- \frac{(V_1-V_2)R_1}{R_1+R_2}$ $i_2 = \frac{V_-}{R_3}$ $V_o = i_2(R_3+R_4) = \frac{V_-}{R_3}(R_3+R_4) = \left(V_1- \frac{(V_1-V_2)R_1}{R_1+R_2}\right)\frac{R_3+R_4}{R_3}$ $V_0 = \left(\frac{V_1(R_1+R_2)-V_1R_1+V_2R_1}{R_1+R_2}\right)\left(\frac{R_3+R_4}{R_3}\right)$ $V_o = (V_1R_2 + V_2R_1)\left(\frac{R_3+R_4}{R_3(R_1+R_2)}\right)$ $V_o = \underbrace{\left(\frac{R_2(R_3+R_4)}{R_3(R_1+R_2)}\right)}_x V_1 + \underbrace{\left(\frac{R_1(R_3+R_4)}{R_3(R_1+R_2)}\right)}_y V_2$

$x = \frac{2(3+4)}{3(1+2)} = \frac{14}{9}$ $y = \frac{1(3+4)}{3(1+2)} = \frac{7}{9}$ $x + y = \frac{2(3+4)}{3(1+2)} = \frac{14}{9} + \frac{7}{9} = \frac{21}{9} = \frac{7}{3}$ Therefore, $a = 7$ and $b = 3$ so $a+ b = \boxed{10}$ .

Considering that OPAMP input currents are $0$ , we can calculate the voltage at the "+" input using superposition:

$V^+=\dfrac{R_2}{R_1+R_2}V_1 + \dfrac{R_1}{R_1+R_2}V_2 = \dfrac{2}{3}V_1 + \dfrac{1}{3}V_2$

The "+" input sees the rest of the circuit as a non-inverting OPAMP configuration, so we can write:

$V_o = (1 + \dfrac{R_4}{R_3})V^+ = \dfrac{7}{3}(\dfrac{2}{3}V_1 + \dfrac{1}{3}V_2) = \dfrac{14}{9}V_1 + \dfrac{7}{9}V_2$