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# Addition and Subtraction w/ different denominators

I swear this was suppose to be easy, but it turned out it was really difficult. Can you show me the complete solution for this equations?

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

THANK YOU VERY MUCH! :-)

ok your denominators have to be exactly the same, then you just add or subtract straight across the top part, the numerator

what i would do first is factor out your denominators

then multiply so that your denominators have the same binomials

1 3/x-5 + 7/5-x

3    +    7
x-5       5-x

3 (5-x)      +   7(x-5)
(x-5)(5-x)    (x-5)(5-x)    <--I took the denominator from the left and multiplied it, top and bottom, on the right.  I did the same thing with the other side, take the denominator from the right and multiply it top and bottom on the left.

15 - 3x                  +     7x-35
5x - x2 -25+5x          5x - x2 -25+5x  <--foil the denominators, combine terms next

15 - 3x        +       7x-35
-x2+10x -25           -x2+10x -25   <-- once everything is cleaned up, you have to add the numerators, when you add, the fractions will combine into one.  dont do anything with the denominators when you add or subtract

15 - 3x + 7x-35
-x2+10x -25              <-- combine terms again

4x + -20
-x2+10x -25

look at the answer:  (4x-20)/(-x^2+10x-25)
both terms can be factored: 4x-20=4(x-5) and (-x^2+10x-25)=(x-5)(-x+5)
both terms have a common factor: (x-5)
the answer in simplest form is 4/(-x+5) or 4/(5-x)