## anonymous 5 years ago ${ 1 \over R} = {1 \over R1} + {1 \over R2}$

1. anonymous

Solve the equation for the indicated variable.

2. anonymous

guess it depends on the indicated variable.

3. anonymous

${1 \over R}$ ? lol

4. anonymous

are you solving for $R$ $R_1$ or $R_2$?

5. anonymous

no. i'm sorry. it's R1

6. anonymous

whew. ok

7. anonymous

haha. work with me satellite. :p

8. anonymous

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

9. anonymous

$\frac{1}{R}-\frac{1}{R_2}=\frac{1}{R_1}$

10. anonymous

${1 \over R} - {1 \over R2} = {1 \over R1} ?$

11. anonymous

$\frac{R_2-R}{RR_2}=\frac{1}{R_1}$

12. anonymous

haha. same thing! this is good!

13. anonymous

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

14. anonymous

where did you come up with the last one?

15. anonymous

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

16. anonymous

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

17. anonymous

oh previous one. ok hold on

18. 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

19. anonymous

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

20. anonymous

clear?

21. anonymous

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

22. anonymous

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

23. anonymous

forgot to close your tag

24. 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?

25. anonymous

but i understood

26. anonymous

yes, but don't get married to that formula. to add and subtract fraction it is always $\frac{a}{bb}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}$

27. anonymous

thanks for confusing me more :p

28. anonymous

just in this case the numerators happened to be 1!

29. anonymous

that should really not confuse you much. how do you add $\frac{2}{3}+\frac{5}{7}$? it is just $\frac{2\times 7 + 3\times 5}{3\times 7}$

30. anonymous

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

31. anonymous

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

32. anonymous

in your case it was $\frac{1}{R}-\frac{1}{R_2}=\frac{1\times R_2-1\times R}{R\times R_2}$

33. anonymous

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

34. anonymous

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

35. anonymous

thank you again! so the: $\frac{a}{bb}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}$ applies to all fractions?