Please help, will medal -3x 4 _____ + ______ x2 – 9 2x – 6
get them in terms of a common denominator Start by factoring the left. recall that \(a^2-b^2=(a+b)(a-b)\)
please use these factorizations: \[\begin{gathered} {x^2} - 9 = \left( {x - 3} \right)\left( {x + 3} \right) \hfill \\ 2x - 6 = 2\left( {x - 3} \right) \hfill \\ \end{gathered} \]
also 4/2x-6=2/x-3
oh ok
Then what @Michele_Laino
Using the factorization above, I get: \[\frac{{ - 3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \frac{4}{{2\left( {x - 3} \right)}} = ...?\]
or, after a simplification: \[\frac{{ - 3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \frac{2}{{\left( {x - 3} \right)}} = ...?\]
ok yes I understand that much, but I don't know what to do with the denominator :/
what is the least common multiple?
I've been very stressed and my brain just isn't understanding math lately, please remind me what a least common multiple is...
the least common multiple is: (x-3)(x+3)
but why? I don't understand
OH!! I just remembered thank you
so does that mean you would have to multiple the right side by (x+3)?
since you have to pick common and uncommon factors with the highest exponent
it means that you have to multiply the numerator of the second fraction, by (x+3)
so multiply both the numerator and denominator by (x+3)?
here is: \[\frac{{ - 3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \frac{2}{{\left( {x - 3} \right)}} = \frac{{ - 3x + 2\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]
note that the numerator of the first fraction is unchanged
so, what is: \[\frac{{ - 3x + 2\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = ...?\]
the denominator is x^2-9 right?
yes!
\[x^{2}-9\]
:) and I'm not sure about the numerator...
you have to compute the multiplication first
what is 2(x+3)=...?
2x+6
so it would be -x+6?
that's right!
THANK YOU SO MUCH
Thank you!
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