Question

How do you change the denominator of a fraction to a reqired denominator without changing the value of that fraction?? PLEASE HELP.!!!!!

Answer 1

lets say you have a fraction:\[\frac{a}{b}\]you know if you multiply this by one then its value won't change - correct?

Answer 2

Yes

Answer 3

so now you can multiply it by another fraction - as long as that fraction evaluates to one, e.g.:\[\frac{a}{b}=\frac{a}{b}*\frac{c}{c}=\frac{ac}{bc}\]

Answer 4

ok

Answer 5

that has now changed the denominator without changing the value of the fraction.

Answer 6

yes, but what if my example is change a/a+b so that the denominator becomes a squared - b squared?

Answer 7

so you fraction is this?:\[\frac{a}{a+b}\]

Answer 8

yes

Answer 9

you need to recall that:\[a^2-b^2=(a+b)(a-b)\]have you seen this before?

Answer 10

yes! it says in my math book that, so i understand that you multiply both the denominator AND the denominator both by a-b

Answer 11

perfect!
you are a pro now :)

Answer 12

alright, so what about a problem like this? ab/3-b: change the denominator to bsquared-9 without changing the value of the fraction?

Answer 13

ok, so you are given:\[\frac{ab}{3-b}\]and need to change it to:\[\frac{something}{b^2-9}\]

Answer 14

yes!

Answer 15

use the same principal:\[b^2-9=(b+3)(b-3)\]in this case there is a small wrinkle in that the denominator contains (3-b) instead of (b-3). but that is easy to resolve as:\[(3-b)=-(b-3)\]can you se what to do now?

Answer 16

Oh! so you factor the given denominator!

Answer 17

not quite factor but rearrange (3-b) into -(b-3)

Answer 18

wait, i dont understand

Answer 19

\[\frac{ab}{3-b}=-\frac{ab}{b-3}\]do you understand this?

Answer 20

yes

Answer 21

ok, so know just multiply numerator and denominator by (b+3) and you are done.

Answer 22

wait i dont understand that wrinkle in the denominator, why did u add a minus sign in front of the b-3?

Answer 23

\[\frac{ab}{3-b}=-\frac{ab}{b-3}=-\frac{ab(b+3)}{(b-3)(b+3)}=-\frac{ab(b+3)}{b^2-9}\]

Answer 24

3 - b = -(b-3)

Answer 25

\[\frac{ab}{3-b}=\frac{ab}{-(b-3)}=-\frac{ab}{b-3}\]

Answer 26

do u have to change it to get the b in front? cant you just use 3-b and multiply that by something to get bsquared -9

Answer 27

you could do this:\[(3-b)(3+b)=9-b^2\]but then the denominator would be \(9-b^2\) instead of \(b^2-9\)

Answer 28

\[\frac{ab}{3-b}=\frac{ab(3+b)}{(3-b)(3+b)}=\frac{ab(3+b)}{9-b^2}\]

Answer 29

so for the final answer's numerator, can you multiply ab by (3+b)?

Answer 30

what do you mean?

Answer 31

the answer can be left as:\[\frac{ab(3+b)}{9-b^2}\]or if you were asked to expand the numerator, then you would get:\[\frac{ab(3+b)}{9-b^2}=\frac{3ab+ab^2}{9-b^2}\]

Answer 32

can u explain the whole precess again?

Answer 33

a lot of these types of problems will be easier if you learn to recognise when to use:\[a^2-b^2=(a+b)(a-b)\]

Answer 34

ok

Answer 35

which part are you confused on?

Answer 36

so u said that b^2-9=(b-3)(b+3) I get that. but how would you solce ab/3-b, when there's no 3-b in that factorization of b^2-9?

Answer 37

(3-b) is just the negative of (b-3). so:\[b^2-9=(b-3)(b+3)=-(3-b)(b+3)=-(3-b)(3+b)\]

Answer 38

oh, so you turn the 3-b into -(3-b) so that you can turn it into a positive!, then what?

Answer 39

no 3-b is not -(3-b).
3-b is -(b-3)

Answer 40

oh whoops, typo

Answer 41

so now you have a (b-3) in the denominator and you can multiply numerator and denominator by (b+3) in order to get the desired \(b^2-9\) in the denominator.

Answer 42

OOOOOOOOOO! so you would change 3-b into (-b+3) to suit the problem, then multiply both top and bottom by (b+3)!!!!!!!! to get ab(b+3)/b^2-9!!!!!!!! can you multiply ab times b+3 on the final answer?

Answer 43

you can expand the numerator by multiplying ab by b+3 but that is not usually required

Answer 44

ok! thank you so much!!!!!!!!!

Answer 45

and again, you would change 3-b into -(b-3) NOT (-b+3) and you typed above

Answer 46

but isn't that the same thing because -(b-3) = -1(b-3) which equals (-b+3)

Answer 47

yes but we were trying to get a term involving (b-3)

Answer 48

ok i understand. thanks!

Answer 49

yw - I'm glad we got there in the end :)

Answer 50

:D