Dividing a Rational Expression

S

stephanie953

New member
Joined
Dec 14, 2006
Messages
40
Hi all. I was helping a friend with her homework and it has been a while since I looked at this stuff. I got stuck on one and it's really bothering me. It is:
a+b/ab / 3ab^2/a-2

So I did this:

a+b/ab / a-2/3ab^2

(a+b)(a-2)/(ab)(3ab^2)

a^2 - 2a + ab -2b/3ab^3

(a^2 + ab)(-2a - 2b)/3ab^3

a(a + b) -2(a + b) /3ab^3

a - 2/3ab^3

I'm sure this is horribly wrong, but I thought I would atleast put what track my mind was on. Any help is greatly appreciated. Thank you.
 
a+b/ab / a-2/3ab^2 means \(\displaystyle a+\frac{b}{a}\cdot \frac{b}{a}-\frac{2}{3}ab^2\). Maybe you could use parenthesis to state what you mean.
I'm guessing you mean ((a+b)/(ab)) /( a-2)/(3ab^2) which means \(\displaystyle \frac{\frac{a+b}{ab}}{\frac{a-2}{3ab^2}}\).
I want to be sure before I waste time working on a problem that is not what is meant.
 
(a+b)/(ab) / (3ab^2)/(a-2)

You have it right except the a-2 should be reversed with the 3ab^2

Thanks
 
\(\displaystyle \frac{\frac{a+b}{ab}}{\frac{3ab^2}{a-2}}\).

Multiply both numerator nd denominator by the least common denominator of the two fractions.

\(\displaystyle \frac{\frac{a+b}{ab}}{\frac{3ab^2}{a-2}} \cdot \frac{\frac{ab(a-2)}{1}}{\frac{ab(a-2)}{1}}\).

Can you take it from there?
 
\(\displaystyle \frac{\frac{a+b}{ab}}{\frac{3ab^2}{a-2}}\)

converting the "major" division to multiplication (of the inverse) - we get

\(\displaystyle = \frac{a+b}{ab} \cdot \frac{a-2}{3ab^2}\)

and continue....
 
This is not looking right to me, but do you end up with:

(a+b)(a-2)
3ab^2(ab)

I know I'm doing something wrong here.

I did (a-2)(ab) a+b/ab + 3ab^2/a-2 (a-2)(ab)

Cancelled out the denominators then had

(a+b)(a-2) * 1/3ab^2(ab)
 
stephanie953 said:
This is not looking right to me, but do you end up with:

(a+b)(a-2)
3ab^2(ab)

I know I'm doing something wrong here.

I did (a-2)(ab) a+b/ab + 3ab^2/a-2 (a-2)(ab)

Cancelled out the denominators then had

(a+b)(a-2) * 1/3ab^2(ab)
Click to expand...

Unless you started with a wrong original problem, you are okay as far as you have gone. You can simplify the denominator by multiplying. You also need to master the use of parenthesis.
(a+b)(a-2) * 1/3ab^2(ab) means \(\displaystyle \frac{(a+b)(a-2)\cdot 1}{3}\cdot ab^2(ab)\)
You should have typed in (a+b)(a-2) * 1/(3ab^2(ab)). Possibly you should study the Order of Operations Convention.
 
I think I wrote the equation down wrong and it was originally how you had said with the a-2 over the 3ab^2. Would that leave me with an a-2 on the bottom of the final answer. Sorry I am trying to help out a friend and I don't have her book. But I am pretty sure the final answer had a-2 in the denominator for the final answer.
 
stephanie953 said:
I think I wrote the equation down wrong and it was originally how you had said with the a-2 over the 3ab^2. Would that leave me with an a-2 on the bottom of the final answer. Sorry I am trying to help out a friend and I don't have her book. But I am pretty sure the final answer had a-2 in the denominator for the final answer.
Click to expand...

[(a + b) / (ab)] / [(a - 2) / (3ab^2)]

= [(a + b) / (ab)] * [(3ab^2) /(a - 2)]

= [(a + b) * (3ab^2)] / [(ab) * (a - 2)]

= [(a + b) * (3b)] / (a - 2) : there you go, ye olde (a - 2) in the detonator :!:
 
Thanks Denis. You're awesome. That was driving me insane. I couldn't figure out why I couldn't get the a-2 denominator.
 
You must log in or register to reply here.
Top