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?
Last edited by mr fantastic; May 1st 2011 at 07:51 PM. Reason: Re-titled.
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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? $\displaystyle \frac{4}{3a} + \frac{3}{3a + 1} = -2$ $\displaystyle \frac{4}{3a} \cdot (3a)(3a + 1) + \frac{3}{3a + 1} \cdot (3a)(3a + 1) = -2 \cdot (3a)(3a + 1)$ $\displaystyle 4(3a + 1) + 3(3a) = -6a(3a + 1)$ Can you finish from here? -Dan
Last edited by topsquark; May 1st 2011 at 05:15 PM.
okay then I just cancel out factors? like the (3a+1) in the numerator and denominator in the first part?
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
okay i got thus far: 7/(3a)(3a+1)=-6a/(3a) how would I solve for a though?
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. $\displaystyle 4(3a + 1) + 3(3a) = -6a(3a + 1)$ $\displaystyle 12a + 4 + 9a = -18a^2 - 6a$ $\displaystyle 18a^2 + 27a + 4 = 0$ -Dan
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