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Mathematics 12 Online
OpenStudy (anonymous):

Simplify this

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@Chlorophyll

OpenStudy (anonymous):

So common denominator is ...?

OpenStudy (anonymous):

Is good.

OpenStudy (anonymous):

Start doing Cross Product now: ...

OpenStudy (anonymous):

I did

OpenStudy (anonymous):

LCD=(a-1)(a+3)

OpenStudy (anonymous):

I got \[a ^{2} + 3 a\] and \[2a ^{2}\]

OpenStudy (anonymous):

2a square - 2a *****

OpenStudy (anonymous):

Show me your multiplication step:

OpenStudy (anonymous):

Okie, correct !

OpenStudy (anonymous):

a x a+3 = \[a ^{2} + 3a\]

OpenStudy (anonymous):

and yeah, the other one.

OpenStudy (anonymous):

Now, add them!

OpenStudy (anonymous):

\[a ^{2} + 3a + 2a ^{2} -2a = 3a ^{2} + a\]

OpenStudy (anonymous):

Correct! Now, plug the common denominator to it!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Although that's not in the answers.

OpenStudy (anonymous):

Some teacher prefer: a ( 3a + 1) in the numerator!

OpenStudy (anonymous):

OH!

OpenStudy (anonymous):

Then it's there.

OpenStudy (anonymous):

Thanks :)

OpenStudy (anonymous):

No, the denominator is the PRODUCT! ( NOT SUM)

OpenStudy (anonymous):

I know.

OpenStudy (anonymous):

It's a huge typo, you know :(

OpenStudy (anonymous):

So, everything's fine now?

OpenStudy (anonymous):

For this part :P

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