anonymous
  • anonymous
\[{ 1 \over R} = {1 \over R1} + {1 \over R2} \]
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Solve the equation for the indicated variable.
anonymous
  • anonymous
guess it depends on the indicated variable.
anonymous
  • anonymous
\[{1 \over R}\] ? lol

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anonymous
  • anonymous
are you solving for \[R\] \[R_1\] or \[R_2\]?
anonymous
  • anonymous
no. i'm sorry. it's R1
anonymous
  • anonymous
whew. ok
anonymous
  • anonymous
haha. work with me satellite. :p
anonymous
  • anonymous
subtract \[\frac{1}{R_2}\] from both sides, take the reciprocal
anonymous
  • anonymous
\[\frac{1}{R}-\frac{1}{R_2}=\frac{1}{R_1}\]
anonymous
  • anonymous
\[{1 \over R} - {1 \over R2} = {1 \over R1} ? \]
anonymous
  • anonymous
\[\frac{R_2-R}{RR_2}=\frac{1}{R_1}\]
anonymous
  • anonymous
haha. same thing! this is good!
anonymous
  • anonymous
now flip everything to get \[\frac{RR_2}{R_2-R}=R_1\]
anonymous
  • anonymous
where did you come up with the last one?
anonymous
  • anonymous
the one before the last... not flipping but the previous one
anonymous
  • anonymous
oh i had 1/R and wanted R. just flip it
anonymous
  • anonymous
oh previous one. ok hold on
anonymous
  • anonymous
\[\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}\] yes? common denominator is ab and i have to build up the fractions
anonymous
  • anonymous
then you subtract to get \[\frac{b-a}{ab}\]
anonymous
  • anonymous
clear?
anonymous
  • anonymous
no. give me a second. i'm going to write it...
anonymous
  • anonymous
ok if not clear let me know. think of \[\frac{1}{5}-\frac{1}{3}\]
anonymous
  • anonymous
forgot to close your tag
anonymous
  • anonymous
this: \[\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab} \] is important for me to remember. i am taking note of it. i think it's just that i need to understand this before being able to do these kind of problems. does that sound right from what you are seeing with me?
anonymous
  • anonymous
but i understood
anonymous
  • anonymous
yes, but don't get married to that formula. to add and subtract fraction it is always \[\frac{a}{bb}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}\]
anonymous
  • anonymous
thanks for confusing me more :p
anonymous
  • anonymous
just in this case the numerators happened to be 1!
anonymous
  • anonymous
that should really not confuse you much. how do you add \[\frac{2}{3}+\frac{5}{7}\]? it is just \[\frac{2\times 7 + 3\times 5}{3\times 7}\]
anonymous
  • anonymous
i would rather find the common denominator for the fraction with numbers lol
anonymous
  • anonymous
but you are showing me a more advanced way. i believe i must learn this way to be succesful
anonymous
  • anonymous
in your case it was \[\frac{1}{R}-\frac{1}{R_2}=\frac{1\times R_2-1\times R}{R\times R_2}\]
anonymous
  • anonymous
so the 1's were eliminated.... then we are left with variables
anonymous
  • anonymous
wtf. that's the easiest way to add fractions that i have seen.
anonymous
  • anonymous
thank you again! so the: \[\frac{a}{bb}\pm\frac{c}{d}=\frac{ad\pm bc}{bd} \] applies to all fractions?

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