http://puu.sh/3KdsI.jpg

hello, i cannot do 10 b and c, the image above shows how much ive gotten done, for b) im not sure how to even start or how i can work through this problem; never seen it before

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- Jul 23rd 2013, 05:32 PMMathnood768rational expressions #3
http://puu.sh/3KdsI.jpg

hello, i cannot do 10 b and c, the image above shows how much ive gotten done, for b) im not sure how to even start or how i can work through this problem; never seen it before - Jul 23rd 2013, 06:27 PMchiroRe: rational expressions #3
Hey Mathnood768.

Hint: Try expanding everything else so that you have expression in the numerator and one in the denominator.

Once you have that, look for a way to factorize expressions and cancel things out. - Jul 23rd 2013, 06:57 PMMathnood768Re: rational expressions #3
yeah i dont know how i would expand it, i know thats what i should do for c) but i dont know what it would be

- Jul 23rd 2013, 07:04 PMchiroRe: rational expressions #3
Hint: a/b + c/d = (ad + bc) / bd

- Jul 23rd 2013, 07:18 PMMathnood768Re: rational expressions #3
can you just show me with my question???

- Jul 23rd 2013, 10:55 PMchiroRe: rational expressions #3
Lets look at b)

2x/(x^3 +x^2 - 6x) - (x-8)/(x^2 - 5x - 24)

= 2x/[x*(x^2 + x - 6)] - (x-8)/[(x-8)(x+3)]

= 2/[x^2 + x - 6] - 1/(x+3)

= 2/[(x+3)(x-2)] - 1/(x+3)

= 2/[(x+3)(x-2)] - (x-2)/[(x-2)(x+3)]

= (4 - x)/[(x+3)(x-2)] (since 2 - x + 2 = 4 - x). - Jul 23rd 2013, 11:32 PMMathnood768Re: rational expressions #3
the answer to b is 3(t+6) / 2(t - 3)

how do i get that - Jul 24th 2013, 02:14 AMProve ItRe: rational expressions #3
Deal with the "biggest" fraction by working on its numerator, then working on its denominator.

How could you simplify $\displaystyle \displaystyle \begin{align*} \frac{3}{2} + \frac{3}{t} \end{align*}$? How would you simplify $\displaystyle \displaystyle \begin{align*} \frac{t}{t + 6} - \frac{1}{t} \end{align*}$? - Jul 24th 2013, 11:30 AMMathnood768Re: rational expressions #3
(3t)(6) / 2t and (t^2)(t+1)/t^2 + 6

??? im not sure - Jul 24th 2013, 07:16 PMProve ItRe: rational expressions #3
It's not $\displaystyle \displaystyle \begin{align*} \frac{3t(6)}{2t} \end{align*}$, I can see you're trying to apply a rule $\displaystyle \displaystyle \begin{align*} \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \end{align*}$ without really understanding what's going on (and messing up the numerator).

To add two fractions, you need a common denominator. In the first your denominator is 2, and in the second your denominator is t. So the common denominator is 2t. Whatever you do to the bottom of each fraction to get the common denominator, you need to do to the top as well. You're really multiplying each fraction by a cleverly disguised 1, because when you multiply by 1, your number remains the same. So

$\displaystyle \displaystyle \begin{align*} \frac{3}{2} + \frac{3}{t} &= \frac{3}{2} \cdot \frac{t}{t} + \frac{3}{t} \cdot \frac{2}{2} \textrm{ (do you see how each of these fractions I've multiplied by is really 1?)} \\ &= \frac{3t}{2t} + \frac{6}{2t} \\ &= \frac{3t + 6}{2t} \end{align*}$ - Jul 24th 2013, 07:37 PMMathnood768Re: rational expressions #3
i see, so for the bottom would it be

t - 1 / t + 6

? - Jul 24th 2013, 07:46 PMProve ItRe: rational expressions #3
No, t+6 is not a common denominator. You will need to use t(t+6).

- Jul 24th 2013, 08:40 PMMathnood768Re: rational expressions #3
(3t + 6 / 2t) / (t -1)/ t(t+6)

idk what to do after this? - Jul 24th 2013, 08:47 PMchiroRe: rational expressions #3
Try using the identity (a/b) / (c/d) = (a*d) / (b*c).

- Jul 24th 2013, 09:47 PMMathnood768Re: rational expressions #3
http://puu.sh/3LbCC.jpg

can i do this?