Simplifying fractions.

**thedroid** · May 1st 2011, 04:46 PM

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?

**topsquark** · May 1st 2011, 05:04 PM

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

**thedroid** · May 1st 2011, 05:11 PM

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

**topsquark** · May 1st 2011, 05:14 PM

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

**thedroid** · May 1st 2011, 05:19 PM

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

**topsquark** · May 1st 2011, 05:21 PM

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

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