## anonymous one year ago 1/a+1/b=1/c for c

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1. anonymous

How do I solve the formula for the specified variable?

2. Astrophysics

$\frac{ 1 }{ a }+\frac{ 1 }{ b }=\frac{ 1 }{ c }$ this?

3. 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

4. anonymous

Do I do c times 1/a ?

5. anonymous

What do you mean divide both sides? Both sides by c?

6. Astrophysics

Well what I mean is we can just divide both sides by $\left( \frac{ 1 }{ a }+\frac{ 1 }{ b } \right)$ so we have $\huge c = \frac{ 1 }{ \left( \frac{ 1 }{ a } + \frac{ 1 }{ b } \right) }$

7. Astrophysics

You can continue to simplify that if you like by using this for the denominator $\huge \frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }$

8. anonymous

If I simplify with the last equation you gave me, where do the C and D variable substitutions come from

9. 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

10. anonymous

1a +/- 1a -------------- ab

11. 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 $\huge c= \frac{ 1 }{ \left( \frac{ b+a }{ ab } \right) }$

12. Astrophysics

Now we just apply that little rule I showed you on your previous question to finish it off

13. anonymous

I have no idea how to simplify this >.<

14. 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 $c = \frac{ ab }{ b+a }$