1. ## Simplifying fractions.

so if i have the equation (4/3a)+[3/(3a+1)]=-2 should my LCD be 3a(3a+1) or just 3a+1? if so how would I multiply it out?

2. Originally Posted by thedroid
so if i have the equation (4/3a)+[3/(3a+1)]=-2 should my LCD be 3a(3a+1) or just 3a+1? if so how would I multiply it out?
$\frac{4}{3a} + \frac{3}{3a + 1} = -2$

$\frac{4}{3a} \cdot (3a)(3a + 1) + \frac{3}{3a + 1} \cdot (3a)(3a + 1) = -2 \cdot (3a)(3a + 1)$

$4(3a + 1) + 3(3a) = -6a(3a + 1)$

Can you finish from here?

-Dan

3. okay then I just cancel out factors? like the (3a+1) in the numerator and denominator in the first part?

4. Originally Posted by thedroid
okay then I just cancel out factors? like the (3a+1) in the numerator and denominator in the first part?
I'm an idiot.

I have corrected my previous post. I plead temporary insanity.

-Dan

5. okay i got thus far: 7/(3a)(3a+1)=-6a/(3a) how would I solve for a though?

6. Originally Posted by thedroid
okay i got thus far: 7/(3a)(3a+1)=-6a/(3a) how would I solve for a though?
I don't know if you've had a chance to see my correction yet.
$4(3a + 1) + 3(3a) = -6a(3a + 1)$

$12a + 4 + 9a = -18a^2 - 6a$

$18a^2 + 27a + 4 = 0$

-Dan