Electrical – Op-amp voltage divider

What I recommend doing is finding common denominator before plugging in, like this:

Node A

$$\frac{V_1-V_{ref}}{R_2} + \frac{V_1-V_{x}}{R_1} + \frac{V_1-V_{2}}{R_f} = 0 $$ $$ V_1(R_1R_2 + R_1R_f + R_2R_f) - V_2R_1R_2 - V_x(R_fR_2) + V_{ref}R_fR_1 = 0$$ $$V_x(R_fR_2) = V_1(R_1R_2 + R_1R_f + R_2R_f) - V_2R_1R_2 + V_{ref}R_fR_1$$

Node B

$$ \frac{V_2-V_{x}}{R_3} + \frac{V_2-V_{1}}{R_f} + \frac{V_2-V_{out}}{R_4} = 0 $$ $$ V_2(R_3R_4 + R_4R_f + R_3R_f) - V_1R_3R_4 - V_x(R_fR_4) + V_oR_fR_3 = 0$$ $$ V_x(R_fR_4) = V_2(R_3R_4 + R_4R_f + R_3R_f) - V_1R_3R_4 + V_oR_fR_3$$

since $$R_4=R_2 \space and \space R_3=R_1$$ $$= V_x(R_fR_2) = V_2(R_1R_2 + R_2R_f + R_1R_f) - V_1R_1R_2 + V_oR_fR_1$$

Then plug it in: $$V_1(R_1R_2 + R_1R_f + R_2R_f) - V_2R_1R_2 + V_{ref}R_fR_1 = V_2(R_1R_2 + R_2R_f + R_1R_f) - V_1R_1R_2 + V_oR_fR_1$$ $$V_oR_fR_1 = V_1(R_1R_2 + R_1R_f + R_2R_f) + V_1R_1R_2 - V_2(R_1R_2 + R_2R_f + R_1R_f) - V_2R_1R_2 + V_{ref}R_fR_1$$ I'm sure you can figure out the rest.

The way I would approach this is to observe that Rf/R3 = 2 (by inspection)

The gain from the non-inverting op-amp input is thus 2 + 1 = 3.

So you need a divider with gain 1/3 from V1 and 2/3 from V2.

Set R1||R2||R4 = R

Then R/R2 = 2/3 and R/R1 = 1/3

Since 1/R4 = 1/R - 1/R2 - 1/R1 = 1/R - 2/3R - 1/3R = 0 So R4 = \$\infty\$

You can pick two more values arbitrarily- say R = 10K which leads to R2 = 15K, R1 = 30K

It would make sense to constrain RF||R3 = 10K to balance the offset from bias currents, so RF = 30K R3 = 15K.

Edit:-

General principle for a voltage divider with n resistors R1...Rn

Vout/Vin = R * (V1/R1+V2/R2+...+VN/RN)

Where R = R1||R2||..||RN

(It's easy to prove this by superposition)