Algebra ??????

[fit1]

Can anyone solve this damn problem????

Solve for b:

1/a + 2/b = 3/c

 

[The Shadow]

Can anyone solve this damn problem????

Solve for b:

1/a + 2/b = 3/c

2

 

[fit1]

okay...uh, i needed to know how to do it corn...lol.
how'd ya' get 2??

 

[mdd]

got me....... lol i'm in college and we're adding fractions right now...... LMAO

 

[The Shadow]

okay...uh, i needed to know how to do it corn...lol.
how'd ya' get 2??

Well - I figured that if a=1, b=2, and c=3, then the quation would be correct. Don't know if that's good logic though...

 

[The Shadow]

Well - I figured that if a=1, b=2, and c=3, then the quation would be correct. Don't know if that's good logic though...

Clear out the fractions:

BC + 2AC = 3AB

BC = 3AB - 2AC

B = 3AB - 2AC
--------------
C

 

[Night Fly]

Hmmm....you aren't solving for a number for B...your answer will be in an equation form with a, b, and c still in it.

I have gotten this far...bear with me....it's been 8 years since algebra.

2/b=3/c + 1/a

Gimme a minute

 

[fit1]

Night....

You got all the time in the world, cause your guess would be better than mine...lol.

Thx Corn...don't know if you're right, but thx...lol

 

[McBainXVII]

ok fit1 now i am certain that i had it right, dot has got the same answer as me, dont ask why i can post in the thread again

 

[Warik]

Days of the Tantric's first answer is right. =)

-Warik

 

[spongebob]

Just substitute as follows and you'll see my answer is correct in it's form:

a=4
b=8
c=6

1/4+2/8=3/6

Then substitute those numbers in my answer.

1/4+2/8=1/2


1/a+2/b=3/c
2/b=3/c-1/a
2/b=3a-c/ac just multiply both sides by 1/2
b=3a-c/2ac

you will not know what the numbers are in this equation.

 

[Warik]

you will not know what the numbers are in this equation.

Actually there are an infinite number of combinations of possible values for a,b, and c. There's no way to get a definate answer unless another equation is given (like for example, a + b = c or a + b = 3 or something)

Heck, I could do:

a = 2
b = 2
c = 1

or

a = 4
b = 4
c = 2

a = 8
b = 8
c = 4

and go on like that until I'm 1,000 years old.

I think the problem just asks to solve for b in terms of a and c. There's really nothing else one could do with such little information.

-Warik

 

[spongebob]

Actually there are an infinite number of combinations of possible values for a,b, and c. There's no way to get a definate answer unless another equation is given (like for example, a + b = c or a + b = 3 or something)

Heck, I could do:

a = 2
b = 2
c = 1

or

a = 4
b = 4
c = 2

a = 8
b = 8
c = 4

and go on like that until I'm 1,000 years old.

I think the problem just asks to solve for b in terms of a and c. There's really nothing else one could do with such little information.

-Warik

i was just checking my answer and its wrong, DOTs is right.
an actually the numbers you gave do not fit into the equation at all,so it might take you a 1000yrs. j/k
heres one, a=2, b=4, c=3