anonymous
  • anonymous
1/a+1/b=1/c for c
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
How do I solve the formula for the specified variable?
Astrophysics
  • Astrophysics
\[\frac{ 1 }{ a }+\frac{ 1 }{ b }=\frac{ 1 }{ c }\] this?
Astrophysics
  • Astrophysics
Few ways, but you can start off by getting the c in the numerator on the left side by multiplying both sides by c. \[c \times \left( \frac{ 1 }{ a } +\frac{ 1 }{ b }\right)=\frac{ 1 }{ c } \times c\] \[c \times \left( \frac{ 1 }{ a } +\frac{ 1 }{ b }\right)= 1\] see if you can finish it off

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anonymous
  • anonymous
Do I do c times 1/a ?
anonymous
  • anonymous
What do you mean divide both sides? Both sides by c?
Astrophysics
  • Astrophysics
Well what I mean is we can just divide both sides by \[\left( \frac{ 1 }{ a }+\frac{ 1 }{ b } \right)\] so we have \[\huge c = \frac{ 1 }{ \left( \frac{ 1 }{ a } + \frac{ 1 }{ b } \right) }\]
Astrophysics
  • Astrophysics
You can continue to simplify that if you like by using this for the denominator \[\huge \frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]
anonymous
  • anonymous
If I simplify with the last equation you gave me, where do the C and D variable substitutions come from
Astrophysics
  • Astrophysics
Those I just used to specify two different fractions so you can see it easier \[\frac{ a }{ b } \implies \frac{ 1 }{ a }\]\[\frac{ c }{ d } \implies \frac{ 1 }{ b }\] don't let all the variables confuse you, it's just place holder from the "equation" above I could have very well used x/y+z/t
anonymous
  • anonymous
1a +/- 1a -------------- ab
Astrophysics
  • Astrophysics
Oh, don't add the plus/ minus sign, I was using that to suggest it's the same whether it's + or -, you can use either equation so you have \[\huge c= \frac{ 1 }{ \left( \frac{ b+a }{ ab } \right) }\]
Astrophysics
  • Astrophysics
Now we just apply that little rule I showed you on your previous question to finish it off
anonymous
  • anonymous
I have no idea how to simplify this >.<
Astrophysics
  • Astrophysics
It's ok, it takes practice to figure out this algebra, it can get confusing with the many rules so we have \[\large \frac{ 1 }{ \left( \frac{ a }{ b } \right) } = \frac{ b }{ a }\] so this implies \[c = \frac{ ab }{ b+a }\]

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