Adding and Subtracting Rational Expressions?

Question

hey can someone help me with this for some reason i can not copy and past my problems into this so they are at the link below please look at the quesion's and post ur answer'e here thanks !!!!!!!!

LINK- http://i143.photobucket.com/albums/r148/dopegurlkeke/08_03_43.gif

Answer 1

1) (5n-4n)/3=n/3

2) (3x+9x)/10=12x/10

3) (n-2+8)/(n+4)=(n+6)/(n+4)

4) (5m+8-m+3)/(m-1)=(4m+11)/(m-1)

5) (x-1)(x+1)/(x-1)=x+1

6) (4x+y-2x+2y)/(2x+3y)=( 2x+3y)/(2x+3y)=1

7) (x^2+3x+2+3x+6)/(x^2-16)= (x^2+6x+8)/(x^2-16)
=(x+2)(x+4)/(x-4)(x+4)
=(x+2)/(x-4)

8) (7-2-x)/(x-5)=(5-x)/(x-5)=-1

Answer 2

you can add or subtract the numerators as long as the denominators are the same

1) 5n/3 - 4n/3 = (5n - 4n)/3 = n/3

2) 3x/10 + 9x/10 = (3x + 9x)/10 = 12x/10 = 6x/5

3) (n-2)/(n+4) + 8/(n+4) = (n - 2 + 8)/(n+4) = (n + 6)/(n+4)

4) (5m+8)/(m-1) - (m-3)/(m-1)
= [ (5m+8)-(m-3)] / (m-1)
= (5m + 8 - m + 3) / (m-1)
= (4m + 11) / (m-1)

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

6) (4x + y)/(2x+3y) - (2x - 2y) / (2x + 3y)
= [ (4x + y) - (2x - 2y) ] / (2x + 3y)
= (4x + y - 2x + 2y) / (2x + 3y)
= (2x + 3y) / (2x + 3y)
= 1

7) (x^2 + 3x + 2) / (x^2 - 16) + (3x + 6) / (x^2 - 16)
= (x^2 + 3x + 2 + 3x + 6) / (x^2 - 16)
= (x^2 + 6x + 8) /(x^2 - 16)
= [(x + 4) ( x + 2) / [(x - 4) (x + 4)]
= (x+2)/(x-4)

8) 7/(x-5) - (2 + x) / (x - 5)
= (7 - (2 + x)) / (x - 5)
= (7 - 2 - x) / (x - 5)
= (5 - x) / (x - 5)
= -(x-5)/(x-5)
= -1

Answer 3

Those are all just fractions with common denominators. Just add or subtract the numerators(top) and put the answer over the same denominator. Ex. 2x/5 +7/5 = 2x+7/5. Your first answer would be 1n/3, or just n/3.