What is the common denominator in this exercise? $\displaystyle \frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{25-9a^2} $ $\displaystyle \frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{(5-3a)(5+3a)} $ Thanks
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hello 1) $\displaystyle \frac{?}{(25-9a^{2})}$ 2) $\displaystyle \frac{?}{(5-3a)(5+3a)}$
So is (5-3a)(5+3a) the common denominator?
Originally Posted by klik11 So is (5-3a)(5+3a) the common denominator? yes
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?
I was thinking about (5-3a)(5+3a)(-5+3a)...
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 $\displaystyle \frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{(5-3a)(5+3a)}=\frac{-a(5+3a)-(a+1)(5-3a)-2(5+a)}{(5-3a)(5+3a) }$ no ?
Originally Posted by Raoh i think we have $\displaystyle \frac{a}{3a-5}-\frac{a+1}{3a+5}-\frac{2(5+a)}{(5-3a)(5+3a)}=\frac{-a(5+3a)-(a+1)(5-3a)-2(5+a)}{(5-3a)(5+3a) }$ no ? Not quite. $\displaystyle (5-3a)\neq(3a-5)$ which means we still need $\displaystyle (3a-5)$ as a factor.
Originally Posted by VonNemo19 Not quite. $\displaystyle (5-3a)\neq(3a-5)$ which means we still need $\displaystyle (3a-5)$ as a factor. yes, but $\displaystyle (5-3a)=-(3a-5)$.
Originally Posted by Raoh yes, but $\displaystyle (5-3a)=-(3a-5)$. Well, in that case... I apologize. Nice work Raoh.
Thanks
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