Intermediate Algebra: Solve 1/a + 1/b = 1/c for c

[Helen]

Could I please get some help with this?

Solve for the specified variable.

1/a + 1/b = 1/c for c
abc * 1/a = abc (1/c + 1/b)
= bc = ab + bc
c= ab/a+b or c = ab(a+b)

Am I putting this together in the right way?

 

[o_O]

Not a bad start but there was a bit of a mess-up in between (although you somehow managed to get the right answer)!

\(\displaystyle \L \frac{1}{a} + \frac{1}{b} = \frac{1}{c}\)

Now we follow what you did and multiply both sides by abc. You didn't need to move 1/b to the otherside but if you did, remember you have to change signs (it should be 1/c - 1/b). However, just leaving everything as it is for now will do:

\(\displaystyle \L abc \cdot \frac{1}{a} + abc \cdot \frac{1}{b} = \frac{1}{c} \cdot abc\)

\(\displaystyle \L bc + ac = ab\)

Now, you have two terms with the common factor c. Can you take it from here?

Also, on your second last line you had bc = ab + bc. If you moved bc to one side, they would cancel and you would get 0 = ab! Which wouldn't work out. Tell us how you're doing when you look at it again :wink:

 

[Denis]

1/a + 1/b = 1/c for c
abc * 1/a = abc (1/c + 1/b)
= bc = ab + bc
c= ab/a+b or c = ab(a+b)
Am I putting this together in the right way?

Much easier if you LCD the left side only:
ab/a + ab/b = 1/c
(b + a) / ab = 1/c
c(b + a) = ab
c = ab / (a + b)

 

[Helen]

Intermediate Algebra

Denis,
Thank you for your help. I will go over it and try it your way.
I am still trying. Helen

 

[Jonathan Plascencia]

It would have been even easier if you were to rewrite the equation to 1/c= 1/a + 1/b. From there you can simply use the butterfly method. You'll get 1/c= (b+a)/(ab). Since youre looking for the value of c and not 1/c, just flip the equation to get c= ab/(b+a), which is the answer. Math is fun because there are always different ways to answer a problem.

 

[topsquark]

Now that's what I call a good necropost! You even got me to thank one of the members!

You are aware that this is a thread from 2007? I have no problems with that, I just thought I'd mention it.

-Dan