Swapping Display Memory on Desk Calculator

v.1
$$\begin{array}{llll} math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\ M=M+D & [\ M+\ ] &B&A+B\\ D=-D & [\ ±\ ] &-B&A+B\\ D=D+M & [\ +\ ]\ [\ MR\ ]\ [\ =\ ] &-B+A+B&A+B\\ M=M-D & [\ M-\ ] &A&A+B-A\\ M ↔ D &done&A&B \end{array}$$

v.2
$$\begin{array}{llll} math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\ D=-D & [\ ±\ ] &-B&A\\ D=D+M & [\ +\ ]\ [\ MR\ ]\ [\ =\ ] &-B+A&A\\ M=M-D & [\ M-\ ] &A-B&A-A+B\\ D="0,=" & [\ 0\ ]\ [\ =\ ] &0(+A)&B\\ M ↔ D &done&A&B \end{array}$$

v.3
$$\begin{array}{llll} math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\ D=D-M & [\ -\ ]\ [\ MR\ ]\ [\ =\ ] &B-A&A\\ M=M+D & [\ M+\ ] &B-A&A+B-A\\ D=D-M & [\ -\ ]\ [\ MR\ ]\ [\ =\ ] &-A+B-B&B\\ D=-D & [\ ×\ ]\ [\ 1\ ]\ [\ -\ ]\ [\ =\ ]\ \left({\ _{[\ =\ ]}\ }\right)^{\ (1} &(-A)·1-((-A)·1)&B\\ M ↔ D &done&A&B \end{array}\\ _{NOTE^{\ (1}\ :\ DEPENDING\ ON\ THE\ SPECIFIC\ CALCULATOR\ THE\ ALTERNATE\ SEQUENCE\\ FOR\ THE\ ARITHMETIC\ NEGATION\ [\ ±\ ]\ MAY\ REQUIRE\ DOUBLE\ [\ =\ ]\ PRESS}$$

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[Eop]

Resistor Bridge

Simulation / Schematic :

$$U_R=\frac{U_LR_2R_4+U_UR_0R_4+U_DR_0R_2}{R_2R_4+R_0R_4+R_0R_2}= \frac{U_L\left({R_1R_3+R_0R_3+R_0R_1}\right)-U_UR_0R_3-U_DR_0R_1}{R_1R_3}$$
$$U_L=\frac{U_RR_1R_3+U_UR_0R_3+U_DR_0R_1}{R_1R_3+R_0R_3+R_0R_1}= \frac{U_R\left({R_2R_4+R_0R_4+R_0R_2}\right)-U_UR_0R_4-U_DR_0R_2}{R_2R_4}$$
$U_RR_1R_3R_2R_4+U_UR_0R_3R_2R_4+U_DR_0R_1R_2R_4=U_R\sum_E^3\sum_O^3-U_UR_0R_4\sum_O^3-U_DR_0R_2\sum_O^3$
$$U_R=R_0·\frac{\left({U_UR_3+U_DR_1}\right)R_2R_4+\left({U_UR_4+U_DR_2}\right)\sum_O^3}{\sum_E^3\sum_O^3-R_1R_2R_3R_4}$$
$U_LR_1R_3R_2R_4+U_UR_0R_4R_1R_3+U_DR_0R_2R_1R_3=U_L\sum_O^3\sum_E^3-U_UR_0R_3\sum_E^3-U_DR_0R_1\sum_E^3$
$$U_L=R_0·\frac{\left({U_UR_4+U_DR_2}\right)R_1R_3+\left({U_UR_3+U_DR_1}\right)\sum_E^3}{\sum_E^3\sum_O^3-R_1R_2R_3R_4}$$

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[Eop]

3 - Resistor Voltage Divider

$$I_0+I_1+I_2=0\qquad,\qquad\text{! Notice the directions of the currents !}$$
$${\large \cases{I_0=\frac{V_0-V_X}{R_0}\\ {\ }\\ I_1=\frac{V_1-V_X}{R_1}\\ {\ }\\ I_2=\frac{V_2-V_X}{R_2}}}$$
$$\frac{V_0-V_X}{R_0}+\frac{V_1-V_X}{R_1}+\frac{V_2-V_X}{R_2}=0$$
$$\frac{V_0R_1R_2-V_XR_1R_2+V_1R_0R_2-V_XR_0R_2+V_2R_0R_1-V_XR_0R_1}{R_0R_1R_2}=0\qquad|\qquad×R_0R_1R_2$$
$$\frac{V_0R_1R_2+V_1R_0R_2+V_2R_0R_1}{R_1R_2+R_0R_2+R_0R_1}=V_X$$
Note: the $V_X$ is the unmarked voltage point on the schematic !

It is easy to see the calculation scheme easily adjusts for N voltages all connecting to the single $V_X$
$$V_{X_N}=\frac{\displaystyle{\sum_{i=0}^{N-1}V_i\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}{\displaystyle{\sum_{i=0}^{N-1}\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}$$
thus BWD chek ::
for N=2 :
$$V_X=\frac{V_1R_0+V_0R_1}{R_0+R_1}$$
incase of the $V_0=0$ it reduces to a trivial 2 resistor divider formula :
$$V_X=V_1\frac{R_0}{R_0+R_1}$$

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Simulation example/chk. for N=4 :

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[Eop]

INA formula chk

$$E_1+\left({V_X-E_1}\right)·\frac{R_2}{R_6+R_2}=E_2+\left({V_R-E_2}\right)·\frac{R_5}{R_7+R_5}$$
$$E_1+\left({E_2-E_1}\right)·\frac{R_1}{R_1+R_4+R_3}=V_1\qquad\frac{E_1-V_1}{R_1}=\frac{E_1-E_2}{R_{\Sigma3}}$$
$$E_2+\left({E_1-E_2}\right)·\frac{R_4}{R_1+R_4+R_3}=V_2\qquad\frac{V_2-E_2}{R_4}=\frac{E_1-E_2}{R_{\Sigma3}}$$
$$E_1=\frac{R_1}{R_4}\left({V_2-E_2}\right)+V_1\qquad E_2=\frac{R_4}{R_1}\left({V_1-E_1}\right)+V_2$$
$$E_2-E_1+\frac{\left({E_1-E_2}\right)R_4+\left({E_1-E_2}\right)R_1}{R_{\Sigma3}}=V_2-V_1$$

$$1-\frac{R_1+R_4}{R_{\Sigma3}}=\frac{V_2-V_1}{E_2-E_1}=\frac{V_2-V_1}{E_2-\frac{R_1}{R_4}\left({V_2-E_2}\right)-V_1}=\frac{V_2-V_1}{\frac{R_4}{R_1}\left({V_1-E_1}\right)+V_2-E_1}$$
$$E_2\left({1+\frac{R_1}{R_4}}\right)-\left({V_2\frac{R_1}{R_4}+V_1}\right)=\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}$$
  $$E_1\left({1+\frac{R_4}{R_1}}\right)-\left({V_2+V_1\frac{R_4}{R_1}}\right)=-\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}$$
$$E_2=\frac{\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}+V_2\frac{R_1}{R_4}+V_1}{1+\frac{R_1}{R_4}}=\frac{\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}R_4+V_2R_1+V_1R_4}{R_1+R_4}$$
$$E_1=\frac{-\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}+V_2+V_1\frac{R_4}{R_1}}{1+\frac{R_4}{R_1}}=\frac{-\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}R_1+V_2R_1+V_1R_4}{R_1+R_4}$$
$$.\ .\ .$$
$$V_X\frac{R_2}{R_6+R_2}-V_R\frac{R_5}{R_7+R_5}=E_2\frac{R_7}{R_7+R_5}-E_1\frac{R_6}{R_6+R_2}$$

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$$IF\ :\ \cases{R_1=R_4=R_A\\ R_3=R_D\\ R_6=R_7=R_F\\ R_2=R_5=R_G}$$
$$E_1=\frac{\cancel{R_A}\left({V_2+V_1}\right)-\frac{V_2-V_1}{1-\frac{2R_A}{2R_A+R_D}}\cancel{R_A}}{2\ \cancel{R_A}}=\frac{V_2+V_1}2-\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}$$
$$E_2=\frac{\cancel{R_A}\left({V_2+V_1}\right)+\frac{V_2-V_1}{1-\frac{2R_A}{2R_A+R_D}}\cancel{R_A}}{2\ \cancel{R_A}}=\frac{V_2+V_1}2+\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}$$
$$.\ .\ .$$

$$\left({V_X-V_R}\right)\frac{R_G}{\cancel{R_F+R_G}}=\left[{\left({\cancel{\frac{V_2+V_1}2}+\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}}\right)-\left({\cancel{\frac{V_2+V_1}2}-\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}}\right)}\right]\frac{R_F}{\cancel{R_F+R_G}}$$
$$\boxed{\frac{V_X-V_R}{V_2-V_1}=\frac{R_F}{R_G}·\frac{2R_A+R_D}{R_D}=\frac{R_6}{R_2}·\frac{2R_1+R_3}{R_3}}$$

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dd

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[Eop]