\[{ 1 \over R} = {1 \over R1} + {1 \over R2} \]

- anonymous

\[{ 1 \over R} = {1 \over R1} + {1 \over R2} \]

- schrodinger

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- anonymous

Solve the equation for the indicated variable.

- anonymous

guess it depends on the indicated variable.

- anonymous

\[{1 \over R}\] ? lol

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## More answers

- anonymous

are you solving for \[R\]
\[R_1\] or \[R_2\]?

- anonymous

no. i'm sorry. it's R1

- anonymous

whew. ok

- anonymous

haha. work with me satellite. :p

- anonymous

subtract \[\frac{1}{R_2}\] from both sides, take the reciprocal

- anonymous

\[\frac{1}{R}-\frac{1}{R_2}=\frac{1}{R_1}\]

- anonymous

\[{1 \over R} - {1 \over R2} = {1 \over R1} ? \]

- anonymous

\[\frac{R_2-R}{RR_2}=\frac{1}{R_1}\]

- anonymous

haha. same thing! this is good!

- anonymous

now flip everything to get
\[\frac{RR_2}{R_2-R}=R_1\]

- anonymous

where did you come up with the last one?

- anonymous

the one before the last... not flipping but the previous one

- anonymous

oh i had 1/R and wanted R. just flip it

- anonymous

oh previous one. ok hold on

- 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

then you subtract to get
\[\frac{b-a}{ab}\]

- anonymous

clear?

- anonymous

no. give me a second. i'm going to write it...

- anonymous

ok if not clear let me know. think of
\[\frac{1}{5}-\frac{1}{3}\]

- anonymous

forgot to close your tag

- 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

but i understood

- 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

thanks for confusing me more :p

- anonymous

just in this case the numerators happened to be 1!

- 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

i would rather find the common denominator for the fraction with numbers lol

- anonymous

but you are showing me a more advanced way. i believe i must learn this way to be succesful

- 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

so the 1's were eliminated.... then we are left with variables

- anonymous

wtf. that's the easiest way to add fractions that i have seen.

- 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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