Question

(a+8)/(a-1) +(a+4)/(a+1) - (8a+2)/(a^2-1)

Answer 1

to solve this , firstly you have to take LCM

Answer 2

your LCM will be a^2-1.

Answer 3

Do you want to end up with a huge expression or a small expression?

Answer 4

I think small.

Answer 5

If you want to end up with a small expression you'll have to take a few steps that are not the norm.

Answer 6

You have

\[\frac{a + 8}{a-1} + \frac{a+4}{a+1} - \frac{8a + 2}{a^2 - 1}\]

Answer 7

And you want to reduce the expression as simply as possible without creating overly large numbers that are annoying to deal with or multiplications that will result in quadratic expressions.

Answer 8

There's a way to avoid such things. What you do is re-write the expressions in the following manner:

\[\frac{a - 1 + 9}{a - 1} + \frac{a + 1 + 3}{a + 1} - \frac{8a + 2}{a^2 - 1}\]

Answer 9

Then split the fractions like so:

\[\frac{a - 1}{a - 1} + \frac{9}{a - 1} + \frac{a + 1}{a + 1} + \frac{3}{a + 1} - \frac{8a + 2}{a^2 - 1}\]

Answer 10

Notice that you can simplify the above so that you have:

\[1 + \frac{9}{a - 1} + 1 + \frac{3}{a + 1} - \frac{8a + 2}{a^2 - 1}\]

and then:

\[2 + \frac{9}{a - 1} + \frac{3}{a + 1} - \frac{8a + 2}{a^2 - 1}\]

Answer 11

I think i should do something like this :
\[\frac{ a+8(a+1) }{ (a+1)(a-1) } ....\]

Answer 12

Yes you could do that....if you want to multiply and get quadratic expressions.

Answer 13

O.K.

Answer 14

But the approach I am taking avoids all that. Watch what happens when I simplify these expressions

Answer 15

I am also going to multiply top and bottom by the appropriate expressions:

Answer 16

\[2 + \frac{9(a + 1)}{(a+1)(a - 1)} + \frac{3(a-1)}{(a-1)(a + 1)} - \frac{8a + 2}{a^2 - 1}\]

Which simplifies to:

\[2 + \frac{9a + 9}{a^2 - 1} + \frac{3a - 3}{a^2 - 1} - \frac{8a + 2}{a^2 - 1}\]

Answer 17

And will also combine the fractions to get:

\[2 + \frac{9a + 9 + 3a - 3 - (8a + 2)}{a^2 - 1} \]
\[2 + \frac{9a + 9 + 3a - 3 - 8a - 2}{a^2 - 1}\]
\[2 + \frac{9a + 3a - 8a + 9 - 3 - 2}{a^2 - 1}\]
\[2 + \frac{(9 + 3 - 8)a + 4}{a^2 - 1}\]
\[2 + \frac{4a + 4}{a^2 - 1}\]
\[2 + \frac{4(a + 1)}{(a+1)(a-1)}\]

Answer 18

Now (a+1)/(a+1) cancels:
\[2 + \frac{4\cancel{(a + 1)}}{\cancel{(a+1)}(a-1)}\]

Leaving just

\[2 + \frac{4}{a-1}\]

Answer 19

We could leave it in that form, as it is the simplest form, however, your instructor might prefer if you express it as one fraction.

Answer 20

So I will multiply 2 by (a - 1)/(a - 1):

\[\frac{2(a -1)}{a-1} + \frac{4}{a-1}\]

This yields:

\[\frac{2a - 2}{a - 1} + \frac{4}{a-1}\]

Which when combined yields:
\[\frac{2a + 4 - 2}{a-1}\]
\[\frac{2a + 2}{a-1}\]

And finally
\[\frac{2(a + 1)}{a-1}\]

Answer 21

Thank you.