Sat prep help

1. Sep 11, 2008

boricua_2004

1/(1/a)-(1/b) a cannot = 0 b cannot equal= 0 and a cannot equal b

i would gladly appreciate if someone can help me

2. Sep 11, 2008

rocomath

Ok ... so what is the problem asking?

3. Sep 11, 2008

boricua_2004

just asking to simplify the problem

4. Sep 11, 2008

rocomath

$$\frac{1}{\frac{1}{a}}-\frac{1}{b}$$

Multiply the numerator of 1/(1/a) by a and the denominator (1/a) by a.

Also, next time post this in the Precalculus section.

5. Sep 11, 2008

boricua_2004

is that wat you think it is or the answer and how did you get there and from my answer choices thats not one them

6. Sep 11, 2008

rocomath

I'm not going to do it for you.

Continue working it with what I suggested with my other reply.

7. Sep 11, 2008

boricua_2004

ok no prob

8. Sep 11, 2008

boricua_2004

ok i have this so far a-1/b im i in the right path

9. Sep 11, 2008

rocomath

Good!

So what is the requirement to add fractions? What do you need in common?

10. Sep 11, 2008

boricua_2004

a denominator if im correct

11. Sep 11, 2008

rocomath

Yep. So a's denominator is just 1, so 1/b determines the denominator needed.

Multiply a's numerator and denominator by what number or letter?

12. Sep 11, 2008

boricua_2004

multiply by b

13. Sep 11, 2008

rocomath

So what do you get?

14. Sep 11, 2008

boricua_2004

b-a please tell me im correct

15. Sep 11, 2008

boricua_2004

because if i multiply by b the b cancels in the denominator and then its a-b or b-a right

16. Sep 11, 2008

rocomath

No, you're only multiplying a by b and writing all of the numerator under a common denominator, b.

Ex:

$$1-\frac{1}{2}=1\cdot\frac{2}{2}-\frac{1}{2}=\frac{2-1}{2}$$

17. Sep 11, 2008

boricua_2004

so it should equal ab-1/b

18. Sep 11, 2008

rocomath

Be careful with how you write or type it.

Is that $$ab-\frac{1}{b}$$ or $$\frac{ab-1}{b}$$???

19. Sep 11, 2008

boricua_2004

ab-1
b

20. Sep 11, 2008

boricua_2004

ab-1/b
the second one