# "Algebric Fractions"?

Question:1) (x+2/x+3)-(x-2/x-3)
2) (5x/x^2-x-6)-(2/x+2)

open the brackets first
1)x+2/x+3-x+2/x+3
x and -x will get cancelled
=2/x+2/x+6
take the LCM
=(2+2+6x^2)/x^2
=4+6x^2/x^2
1.) for the frst one multiply the denominator by x-3 andthe other one by x+3 and by taking lcm solve

2.) factorize the denominator of the frst one and take lcm of the two and solve
1) (x + 2/x + 3) - (x - 2/x -3)
= x + 2/x + 3 - x + 2/x +3
= 2/x +2/x +3
= 4/x + 3

2) (5x/x^2 - x - 6) - (2/x + 2)
=5/x - x - 6 - 2/x + 2
=7/x - x - 4

think this should be correct
I do not get the same answers as the other people did... To add or subtract fractions, you need to have a common denominator.

First problem

1) The common denominator is (x+3)(x-3). Convert both of your fractions so they have this denominator.

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

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

2) Multiply these out.

(x^2 - x - 6)/(x^2 - 9) - (x^2 +x -6)/(x^2 -9)

3) Now combine like terms in the numerator. Everything is eliminated except -2x, so the answer is -2x / (x^2 - 9)

Second problem:

1) Find the common denominator. x^2 - x - 6 = (x+2)(x-3), so since the other denominator is (x+2), you will only need to convert the second term.

5x/(x+2)(x-3) - [2/(x+2) * (x-3)/(x-3)]

2) Multiply this out.
5x/(x+2)(x-3) - 2(x-3)/(x+2)(x-3) =

5x/(denominator) - (2x - 6)/(denominator) =

(3x + 6)/denominator, factored is 3(x +2).

3(x+2)/(x+2)(x-3) can be reduced to 3/(x-3).

Did this help?