Adding/Subtracting/Multiplying/Dividing Rational Expressions [+rep]

[Maki]

Need explanations pls I keep getting the denominator right but I fuck up on the numerator, always always always


a) 1 / x^2-5x+6 - 1 / x^2-9
Ans:


b) 1 / 2x - 7 / 3x + 4 / x^3
Ans:


c) 3x / x+2 + 4x/ x-6
Ans:


d) 6x / x^2-5x+6 - 3x / x^2+x-12
Ans:


Fukin' math s2g
In desperate need of answers + walkthroughs, I'm gonna punch a baby and then punch myself

[Tablo]

Please refrain from punching babies! (I cant help but good luck)

[boomer]

a) 1 / x^2-5x+6 - 1 / x^2-9

Factor the denominators:
x^2-5x+6 = (x-3)(x-2)
x^2-9 = (x-3)(x+3)

There are three unique terms for the denominators, which multiplied together give you your common denominator that you should aim for: (x-3)(x+3)(x-2)
(Also, compare this denominator to the answer!)

The first term only has (x-3)(x-2) in the denominator. To get to the common denominator, multiply both the top and bottom of the first term by (x+3), the missing term.

So now 1 / x^2-5x+6 = (x+3)/(x-3)(x+3)(x-2)

The second term only has (x-3)(x+3). Multiply top and bottom by the missing term (x-2)

1 / x^2-9 = (x-2)/(x-3)(x+3)(x-2)

Now the expression looks like (x+3)/(x-3)(x+3)(x-2) - (x-2)/(x-3)(x+3)(x-2).
Subtracting the numerators gives you x+3-x+2 = 5
The denominator stays the same.
That is your answer.

(Working on typing out the other ones have no fear~)

---------- Post added at 07:34 PM ---------- Previous post was at 07:24 PM ----------

Can you double check that the second term is not actually 7/3x^2? Otherwise you don't get that answer.

b) 1 / 2x - 7 / 3x^2 + 4 / x^3

Again, look for a common denominator.
If you multiply 2x and 3x^, you get 6x^3 which includes x^3 as well.

How can you tell this works? 6x^3 is divisible by every other denominator present.
6x^3/2x = 3x^2, 6x^3/3x = 2x, 6x^3/x^3 = 6

Now you multiple each term (top and bottom) by whatever terms it needs so that the denominator is the common denominator.
First term: 1 / 2x * (3x^2/3x^2) = 3x^2 / 6x^3
Second term: 7 / 3x * (2x/2x) = 14x / 6x^3
Third term: 4 / x^3 * (6/6) = 24 / 6x^3

So, 1 / 2x - 7 / 3x + 4 / x^3 becomes (3x^2 - 14x + 24) / 6x^3 (Answer)

---------- Post added at 07:38 PM ---------- Previous post was at 07:34 PM ----------

c) 3x / x+2 + 4x/ x-6

The common denominator is (x+2)(x-6)

Multiply the first term top and bottom by the missing term, (x-6)
3x / x+2 * (x-6)/(x-6) = 3x(x-6) / (x+2)(x-6)

Multiply the second term the same way by (x+2)
4x/ x-6 * (x+2)/(x+2) = 4x(x+2) / (x+2)(x-6)

Looking at your numerators, multiply them out:
3x^2 -18x + 4x^2 + 8x = 7x^2 - 10x

Put this over the common denominator = Answer

[Doset]

Question a)

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edit: nvm I'll let boomer do the job then :p

[boomer]

d) 6x / x^2-5x+6 - 3x / x^2+x-12

Factor both denominators:
x^2-5x+6 = (x-2)(x-3)
x^2+x-12 = (x-3)(x+4)

Common denominator: (x-2)(x-3)(x+4)

First term:
Missing? x+4
Multiply top and bottom by x+4 so you get
6x(x+4) / (x-2)(x-3)(x+4)

Second term:
Missing? x-2
Multiply top and bottom by x-2 so you get
3x(x-2) / (x-2)(x-3)(x+4)

Multiply out and combine your numerators:
6x(x+4) - 3x(x-2) = 6x^2 + 24x - 3x^2 + 6x = 3x^2 + 30x

Put this over the common denominator = Answer


And in every case, you can't have x be a value that makes your denominator 0 because then the expression is undefined.

General steps:
1. Find the common denominator
2. Multiply each term both top and bottom by terms that will get the denominator to the common denominator
3. Add or subtract up your numerators and put over the common denominator

---------- Post added at 07:44 PM ---------- Previous post was at 07:43 PM ----------

Lol @(you need an account to see links) but you have it in pretty math and here I am with ugly forum math! *lazy*

Hope it helps @(you need an account to see links)

[Maki]

@(you need an account to see links) thanks so much! Just slowly going through each of your explanations so far!

I'm on b), but I can't really follow. Is there any way to do it in such that you would subtract the first two terms, and then add the difference with 4 / x^3?

[Raj]

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seeing it worked out might be easier than the written interpretations?

[boomer]

@(you need an account to see links) thanks so much! Just slowly going through each of your explanations so far!

I'm on b), but I can't really follow. Is there any way to do it in such that you would subtract the first two terms, and then add the difference with 4 / x^3?

Yes, taking this problem in two separate steps:

First 2 terms:

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Putting the last term in:

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[Maki]

Yes, taking this problem in two separate steps:

First 2 terms:

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Putting the last term in:

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Aw boomer don't feel bad about your lazy forum math I actually quite prefer carefully written out explanations!
You saved my life tonite. I will now commence loving u forever
@(you need an account to see links) another baby is spared..for now ;D

Thanks @(you need an account to see links) and @(you need an account to see links), +repped you both too

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[Tablo]

Aw boomer don't feel bad about your lazy forum math I actually quite prefer carefully written out explanations!
You saved my life tonite. I will now commence loving u forever
@(you need an account to see links) another baby is spared..for now ;D

Thanks @(you need an account to see links) and @(you need an account to see links), +repped you both too

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I guess its my duty to rep him for you, he did save that baby after all...