Adding & Subtracting w/ Different denominators:

1. 3/x-5 + 7/5-x

2. 30-4x/x^2-9 + 7/x+3

3. 5x/5x^2-125 + 5x-5/x^2-6x+5

4. 2/x-2 - 10/x^2+x-6

5. 8/6x-18 - 14/6-2x

6. 2/x^2-4 - 5/x^2-3x-10

7. x+1/3x^2-2x-1 + x-1/3x^2+4x+1

8. 2/a - 3/a+1 + 5/a-1

Perform the indicated opertions:

1. 2z^2-3z+6/z^2-1 - z^2-5z+9/z^2-1

2. 3/6x^2-4x - x-2/9x-6

3. 2x+1/6x^2+3x+1 + 2x-1/6x^2-x-1

4. (ab)^2/(a+b)^2 * (a+b)^3/(ab)^3

5. 8a/2a^2+4a+2 - 3a-3/a^2-1

6. 3a^2-2a-16/2a^2+3a-2 * 6a+16/9a^2-64

7. a-3/a^3+8 - 2/a+2 - a-3/a^2-2a+4

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Marked as Best Answer

2) 30-4x/x^2-9 + 7/(x+3)

30-4x/(x-3)(x+3) + 7/(x+3) (I factored x^2-9)

30-4x/(x-3)(x+3) + 7(x-3)/(x+3)(x-3)

30-4x + 7(x-3)/(x+3)(x-3)

30-4x+7x-21/(x+3)(x-3)

30+3x-21/(x+3)(x-3)

9+3x/(x+3)(x-3)

3(3+x)/(x+3)(x-3)

3/(x-3)

4) (ab)^2/(a+b)^2 * (a+b)^3/(ab)^3

simplify

(ab)^2 goes into (ab)^2 1 time, (ab)^2 goes into (ab)^3 (ab) times

so we have so far 1/(a+b)^2 * (a+b)^3/(ab)

now divide the numerator and the denominator by (a+b)^2

1/1 * (a+b)/(ab)

answer: (a+b)/(ab)

5) I'll factor the denominators for you.

2a^2+4a+2=2(a^2+2a+1)=2(a+1)(a+1)

a^2-1=(a+1)(a-1)

6) I'll factor all the terms for you and you finish the problem !

(3a-8)(a+2)/(2a-1)(a+2) * 2(3a+8)/(3a-8)(3a+8)

Simplify. Look at the common factors in the terms.

Hi Page,

Unfortunately I'm kind of pressed for time this morning, however adding and subtracting with different denominators is really a simple process once you get use to the concept. In order to add two different denominators you have to multiply each side of the problem by the denominator of the opposite side where the polynomial/ function is written twice in the numerator and denominator.

So to make sense of that, since it is worded poorly look lets look at problem 1.

3/(x-5) + 7/(5-x)

in order to add these two the next operation you'll need to do is that cross multiply, which looks like this

[3(5-x)]/[(x-5)(5-x)] + [7(x-5)]/[(5-x)(x-5)]

the trick is to feta. Common denominator and this is done through the multiplication

from there, technically this answer is correct, but you will need to simplify by distributing the top then adding the two functions together. Make sure you leave the denominator, typically, as is so that you can see if it won't cancel nicely.

sorry I can't actually do a lot of the problems for you, but if hope you have a better understanding now! Good luck and hopefully I can get back to you on these later this weekend!

Best,

Evan

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