The negative sum is obtained by setting $V_3$ = 0. As shown in Figure 1,

$$V'_O = -\frac{R_F}{R_1}V_1 - \frac{R_F}{R_1}V_2$$
$$V'_O= -\frac{R_F}{R_1}(V_1 + V_2)$$
Given $R_F$ = 100K, $R_1$ = 10K, $V_1$ = 1.5V and $V_2$ = 3V
$$V'_O = -\frac{100K}{10K}(1.5 + 3)$$
$$\boxed{V'_O = -45V}..........(1)$$

Now set $V1$ = $V2$ = 0V, in order to find output voltage due $V3$ ,

Since 10K becomes parallel to 10K as $V_1$ and $V_2$ are grounded hence equivalent resistance R becomes as shown in Figure 2,

$$R = 10K || 10K = \frac{10K X 10K}{10K + 10K} = 5KΩ......(2)$$
Calculate the output voltage due to $V_3$:
$$V^"_O = (1 + \frac{R_F}{R})V'_3.........(3)$$

Where $V'_3$ is the voltage obtained due to Voltage Divider Rule as shown in Figure 3,

$$V'_3 = \frac{1.5K}{10K + 1.5K}V_3 = 0.130 X 4 = 0.52V....(4) $$

Substituting equation (2) and equation (4) in equation (3),
$$V^"_O = (1 + \frac{100K}{5K}) X 0.52 = 10.92V$$
$$\boxed{V^"_O = 10.92 V}.......(5)$$

Adding equation (1) and (5) to obtain total output voltage $V_O$,
$$V_0 = V'_0 + V^"_0$$
$$V_0 = -45 + 10.92$$
$$\boxed{V_O = -34.08 V}$$

## The output voltage of amplifier is $V_O$ = -34.08V