1. 2 a + 1 4
------- + ---------- = -----
a - 3 a^2 - 9 a + 3
2. x 1
----- = -----
x + 2 x
Solutions:::
For example, for the fractions 2/7 and 1/3, you can multiply 7 * 3 to
get 21, one common denominator.
For your first problem, a^2 - 9 will be the lowest common denominator.
[I know this because of the "difference of squares" factoring rule.
Since a^2 - 9 is the difference of squares, we can factor it as
(a - 3)(a+3)]
So for the first problem, we will need to multiply fractions with the
denominator of a-3 by a+3, (which equals 1) and fractions with the
---
a+3
denominator of a+3 by a-3 (also equal to 1). Does this make sense?
---
a-3
2 a + 3 = 2(a+3) = 2a+6
-- * ------ ------ -----
a-3 a + 3 a^2 -9 a^2 -9
We can leave a+1 as it is for the moment.
---
a^2 -9
We will need to multiply
4 a-3 4 (a-3) 4a -12
- * --- to get ------- = ------
a+3 a-3 a^2 -9 a^2 -9
So now the problem reads
2a +6 +a+1 = 4a -12
----------- ------
a^2 -9 a^2 -9
Does that look better?
You can now multiply through by a^2 -9 to remove the denominators if
you like and continue solving this like any other algebra problem.
Are you comfortable with the rest of the problem? If not, write us
back and we'll give you a hand.
Here's just a clue for the second problem. Again, if you want more
help, let us know. For the second problem, the common denominator is
x(x+2) (aka x^2 + 2x)
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