fit1 algebra problem

[McBainXVII]

ok the bastard wouldnt let me post in the last one i think its

b= 2ac/3a-c but its late and im not a genius so it might be wrong

 

[Night Fly]

I'm trying to figure that one out, too...can you put up on here how you did it?

 

[McBainXVII]

1/a + 2/b = 3/c

bc+ 2ac = 3ab

2ac = 3ab - bc

2ac = b(3a-c)

b= 2ac/3a-c

hows that??

 

[fit1]

i know you have to get rid of the fractions first, but HOW do you get rid of them? since they are all different?

 

[Night Fly]

Yeah, can you put in the step showing where you got rid of the fractions?

 

[McBainXVII]

ok to get rid of the fractions you have to multiply each numerator by the other two denominators, sort of like cross multiplying: do that to each numerator and the fractions disappear

 

[The Shadow]

Yeah, can you put in the step showing where you got rid of the fractions?

1/a + 2/b = 3/c

To clear the fractions, multiply by the LCD(least common denominator) which is abc

you are left with:

bc + 2ac = 3ab

bc = 3ab - 2 ac

divide by c to solve for b

b= 3ab-2ac all over c

 

[McBainXVII]

nice one

 

[Night Fly]

Yeah, I think corn got it....

When I worked it out like McBain explained...I got this...

b=(3-2ac)/c

I think the way cornholio did it was the right way...because you can't cross multiply when you are adding...I think I am right on that...but, who knows.

Fit1, you have stumped some people this morning! Good job, girl!

 

[fit1]

A-HA!!!
I get it now..thx Corn and McBain and of course Night...lol
I'll be emailing y'all 2 if i get stumped again...in other words,
give me a minute and let me find another problem...lol..j/k

 

[McBainXVII]

wait!! soryr to be a pain in the ass but do you want an answer for b or an answer for b/c??, youve gotta take out a common factor of b to get b on its own and my computer kjeeps fucking up

 

[fit1]

McBain - I think the way Corn did it was right. After you get rid of the fractions, you get the bc by itself. and then divide by c and that puts b by itself...so you have b=3ab+2ac/c

Night - I know I did...heehee...I finally had a post that someone actually responded to...lol.

 

[doc_pup]

hmm...here's what i got
(1/a)+(2/b)=(3/c)
2/b=3/c-1/a
(b)(2/b)=(3/c-1/a)(b)
2=3b/c-b/a
2=3bc-bc/ac
2=b(3c-c)/ac
(ac)(2)=b(2c)/ac(ac)
2ac=b(2c)
b=2ac/2c
b=ac/c

 

[fit1]

mmm-kay..now i think i'm gettin' confused again...lol

 

[doc_pup]

aha, i got it now, made a little error in common denominator there
2/b=3/c-1/a
2=3ba/ac-bc/ac
2=b(3a-c)/ac
2ac=b(3a-c)
b=2ac/3a-c

i checked it with my graphing calculator and this is what it came up with