Equation simplification

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Sorry if the question is pretty trivial but can anyone tell me what rule or law was used to go from $$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$$ $$to$$ $$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$$




arithmetic




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  • Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly
    – Davide Morgante
    Jul 19 at 16:27














up vote
1
down vote

favorite












Sorry if the question is pretty trivial but can anyone tell me what rule or law was used to go from $$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$$ $$to$$ $$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$$




arithmetic




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  • Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly
    – Davide Morgante
    Jul 19 at 16:27












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Sorry if the question is pretty trivial but can anyone tell me what rule or law was used to go from $$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$$ $$to$$ $$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$$




arithmetic




share|cite|improve this question











Sorry if the question is pretty trivial but can anyone tell me what rule or law was used to go from $$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$$ $$to$$ $$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$$





arithmetic





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asked Jul 19 at 15:52









CptPackage

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  • Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly
    – Davide Morgante
    Jul 19 at 16:27
















  • Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly
    – Davide Morgante
    Jul 19 at 16:27















Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly
– Davide Morgante
Jul 19 at 16:27




Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly
– Davide Morgante
Jul 19 at 16:27










2 Answers
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$$e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4colorred-R_7R_2^-1e_2 \ e_2colorred+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1 \
colorblue(1+R_7R_2^-1)e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4 \
e_2 = fracC_4^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_1 + fracC_10^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_2-fracR_7I_3^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_4 \
e_2=fracC_4^-1R_2R_2+R_7x_1+fracC_10^-1R_2R_2+R_7x_2-fracR_3I_3^-1R_2R_2+R_7x_4$$






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  • Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
    – CptPackage
    Jul 19 at 16:22










  • Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
    – Davide Morgante
    Jul 19 at 16:26











  • Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
    – CptPackage
    Jul 19 at 16:38










  • Ahah! Grazie mille, un saluto dalla Sapienza :P
    – Davide Morgante
    Jul 19 at 16:38

















up vote
1
down vote













$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$



$e_2+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2(1 + R_7R_2^-1)= C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2 = fracC_4^-1x_11 + R_7R_2^-1+fracC_10^-1x_21 + R_7R_2^-1-fracR_7I_3^-1x_41 + R_7R_2^-1$



Multiply each numerator and denominator by $R_2$ to get.........



$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$






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  • Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
    – CptPackage
    Jul 19 at 16:22










  • Having a negative exponent in the numerator is bad form.
    – Phil H
    Jul 19 at 17:56










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2 Answers
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2 Answers
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up vote
1
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accepted










$$e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4colorred-R_7R_2^-1e_2 \ e_2colorred+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1 \
colorblue(1+R_7R_2^-1)e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4 \
e_2 = fracC_4^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_1 + fracC_10^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_2-fracR_7I_3^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_4 \
e_2=fracC_4^-1R_2R_2+R_7x_1+fracC_10^-1R_2R_2+R_7x_2-fracR_3I_3^-1R_2R_2+R_7x_4$$






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  • Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
    – CptPackage
    Jul 19 at 16:22










  • Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
    – Davide Morgante
    Jul 19 at 16:26











  • Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
    – CptPackage
    Jul 19 at 16:38










  • Ahah! Grazie mille, un saluto dalla Sapienza :P
    – Davide Morgante
    Jul 19 at 16:38














up vote
1
down vote



accepted










$$e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4colorred-R_7R_2^-1e_2 \ e_2colorred+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1 \
colorblue(1+R_7R_2^-1)e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4 \
e_2 = fracC_4^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_1 + fracC_10^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_2-fracR_7I_3^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_4 \
e_2=fracC_4^-1R_2R_2+R_7x_1+fracC_10^-1R_2R_2+R_7x_2-fracR_3I_3^-1R_2R_2+R_7x_4$$






share|cite|improve this answer





















  • Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
    – CptPackage
    Jul 19 at 16:22










  • Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
    – Davide Morgante
    Jul 19 at 16:26











  • Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
    – CptPackage
    Jul 19 at 16:38










  • Ahah! Grazie mille, un saluto dalla Sapienza :P
    – Davide Morgante
    Jul 19 at 16:38












up vote
1
down vote



accepted







up vote
1
down vote



accepted






$$e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4colorred-R_7R_2^-1e_2 \ e_2colorred+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1 \
colorblue(1+R_7R_2^-1)e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4 \
e_2 = fracC_4^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_1 + fracC_10^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_2-fracR_7I_3^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_4 \
e_2=fracC_4^-1R_2R_2+R_7x_1+fracC_10^-1R_2R_2+R_7x_2-fracR_3I_3^-1R_2R_2+R_7x_4$$






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$$e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4colorred-R_7R_2^-1e_2 \ e_2colorred+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1 \
colorblue(1+R_7R_2^-1)e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4 \
e_2 = fracC_4^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_1 + fracC_10^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_2-fracR_7I_3^-1colorblue(1+R_7R_2^-1)colorredfracR_2R_2x_4 \
e_2=fracC_4^-1R_2R_2+R_7x_1+fracC_10^-1R_2R_2+R_7x_2-fracR_3I_3^-1R_2R_2+R_7x_4$$







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answered Jul 19 at 16:18









Davide Morgante

1,830220




1,830220











  • Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
    – CptPackage
    Jul 19 at 16:22










  • Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
    – Davide Morgante
    Jul 19 at 16:26











  • Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
    – CptPackage
    Jul 19 at 16:38










  • Ahah! Grazie mille, un saluto dalla Sapienza :P
    – Davide Morgante
    Jul 19 at 16:38
















  • Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
    – CptPackage
    Jul 19 at 16:22










  • Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
    – Davide Morgante
    Jul 19 at 16:26











  • Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
    – CptPackage
    Jul 19 at 16:38










  • Ahah! Grazie mille, un saluto dalla Sapienza :P
    – Davide Morgante
    Jul 19 at 16:38















Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
– CptPackage
Jul 19 at 16:22




Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore?
– CptPackage
Jul 19 at 16:22












Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
– Davide Morgante
Jul 19 at 16:26





Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^-1=fracR_7R_2$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$fracC_4^-11+fracR_7R_2 = fracC_4^-1fracR_2+R_7R_2$$ dopodiché, utilizzando il fatto che $$fracafracbc = fracacb$$ possiamo arrivare alla forma finale $$fracC_4^-1fracR_2+R_7R_2 = fracC_4^-1R_2R_2+R_7$$
– Davide Morgante
Jul 19 at 16:26













Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
– CptPackage
Jul 19 at 16:38




Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta.
– CptPackage
Jul 19 at 16:38












Ahah! Grazie mille, un saluto dalla Sapienza :P
– Davide Morgante
Jul 19 at 16:38




Ahah! Grazie mille, un saluto dalla Sapienza :P
– Davide Morgante
Jul 19 at 16:38










up vote
1
down vote













$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$



$e_2+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2(1 + R_7R_2^-1)= C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2 = fracC_4^-1x_11 + R_7R_2^-1+fracC_10^-1x_21 + R_7R_2^-1-fracR_7I_3^-1x_41 + R_7R_2^-1$



Multiply each numerator and denominator by $R_2$ to get.........



$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$






share|cite|improve this answer





















  • Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
    – CptPackage
    Jul 19 at 16:22










  • Having a negative exponent in the numerator is bad form.
    – Phil H
    Jul 19 at 17:56














up vote
1
down vote













$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$



$e_2+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2(1 + R_7R_2^-1)= C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2 = fracC_4^-1x_11 + R_7R_2^-1+fracC_10^-1x_21 + R_7R_2^-1-fracR_7I_3^-1x_41 + R_7R_2^-1$



Multiply each numerator and denominator by $R_2$ to get.........



$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$






share|cite|improve this answer





















  • Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
    – CptPackage
    Jul 19 at 16:22










  • Having a negative exponent in the numerator is bad form.
    – Phil H
    Jul 19 at 17:56












up vote
1
down vote










up vote
1
down vote









$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$



$e_2+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2(1 + R_7R_2^-1)= C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2 = fracC_4^-1x_11 + R_7R_2^-1+fracC_10^-1x_21 + R_7R_2^-1-fracR_7I_3^-1x_41 + R_7R_2^-1$



Multiply each numerator and denominator by $R_2$ to get.........



$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$






share|cite|improve this answer













$e_2=C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4-R_7R_2^-1e_2$



$e_2+R_7R_2^-1e_2 = C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2(1 + R_7R_2^-1)= C_4^-1x_1+C_10^-1x_2-R_7I_3^-1x_4$



$e_2 = fracC_4^-1x_11 + R_7R_2^-1+fracC_10^-1x_21 + R_7R_2^-1-fracR_7I_3^-1x_41 + R_7R_2^-1$



Multiply each numerator and denominator by $R_2$ to get.........



$e_2=dfracR_2*C_4^-1R_2+R_7x_1+dfracR_2*C_10^-1R_2+R_7x_2-dfracR_2*R_7*I_3^-1R_2+R_7x_4$







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answered Jul 19 at 16:13









Phil H

1,8212311




1,8212311











  • Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
    – CptPackage
    Jul 19 at 16:22










  • Having a negative exponent in the numerator is bad form.
    – Phil H
    Jul 19 at 17:56
















  • Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
    – CptPackage
    Jul 19 at 16:22










  • Having a negative exponent in the numerator is bad form.
    – Phil H
    Jul 19 at 17:56















Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
– CptPackage
Jul 19 at 16:22




Thanks but the reason is why did we multiply numerator and denominator by $R_2$?
– CptPackage
Jul 19 at 16:22












Having a negative exponent in the numerator is bad form.
– Phil H
Jul 19 at 17:56




Having a negative exponent in the numerator is bad form.
– Phil H
Jul 19 at 17:56












 

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