common denominator

**klik11** · November 8th 2009, 07:12 AM

What is the common denominator in this exercise?
$<br /> \frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{25-9a^2}<br />$

$<br /> \frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{(5-3a)(5+3a)}<br />$

Thanks

**Raoh** · November 8th 2009, 07:25 AM

hello

1) $\frac{?}{(25-9a^{2})}$
2) $\frac{?}{(5-3a)(5+3a)}$

**klik11** · November 8th 2009, 07:29 AM

So is (5-3a)(5+3a) the common denominator?

**Raoh** · November 8th 2009, 07:30 AM

Originally Posted by klik11

So is (5-3a)(5+3a) the common denominator?

yes

**VonNemo19** · November 8th 2009, 07:33 AM

Originally Posted by klik11

So is (5-3a)(5+3a) the common denominator?

Nope.

When finding a LCD, forget about the numerators for a second and focus only on the denominators.

In your problem, what factors would be needed in the last term so that it has it contains the factors of the two other terms?

**klik11** · November 8th 2009, 07:38 AM

I was thinking about (5-3a)(5+3a)(-5+3a)...

**Raoh** · November 8th 2009, 07:42 AM

Originally Posted by VonNemo19

Nope.

When finding a LCD, forget about the numerators for a second and focus only on the denominators.

In your problem, what factors would be needed in the last term so that it has it contains the factors of the two other terms?

i think we have $\frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{(5-3a)(5+3a)}=\frac{-a(5+3a)-(a+1)(5-3a)-2(5+a)}{(5-3a)(5+3a) }$
no ?

**VonNemo19** · November 8th 2009, 07:47 AM

Originally Posted by Raoh

i think we have $\frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{(5-3a)(5+3a)}=\frac{-a(5+3a)-(a+1)(5-3a)-2(5+a)}{(5-3a)(5+3a) }$
no ?

Not quite.

$(5-3a)\neq(3a-5)$ which means we still need $(3a-5)$ as a factor.

**Raoh** · November 8th 2009, 07:52 AM

Originally Posted by VonNemo19

Not quite.

$(5-3a)\neq(3a-5)$ which means we still need $(3a-5)$ as a factor.

yes, but $(5-3a)=-(3a-5)$ .

**VonNemo19** · November 8th 2009, 07:55 AM

Originally Posted by Raoh

yes, but $(5-3a)=-(3a-5)$ .

Well, in that case...

I apologize.

Nice work Raoh.

**klik11** · November 8th 2009, 07:57 AM

Thanks

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