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18jonea
 one year ago
find the domain and simplify
18jonea
 one year ago
find the domain and simplify

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18jonea
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ x+1 }{ 3x3 }  \frac{ 4 }{ x^23x+2 }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to factorize both denominators

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1i know and i did the first but dont know how to do the decond first one is 3(x1)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for the second one, you have to search for two numbers such that their sum is 3, and their product is 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we can write this: \[\Large {x^2}  3x + 2 = \left( {x  1} \right)\left( {x  2} \right)\]

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1now we have to find the domain

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2therefore: \[\Large \begin{gathered} \frac{{x + 1}}{{3x  3}}  \frac{4}{{{x^2}  3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x  1} \right)}}  \frac{4}{{\left( {x  1} \right)\left( {x  2} \right)}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! we have to find the domain

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2in order to do that, we note that we can not divide by zero

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1Domain is all real numbers except when x = 1 or 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we have to request that: \[\Large \begin{gathered} x  1 \ne 0 \hfill \\ \left( {x  1} \right)\left( {x  2} \right) \ne 0 \hfill \\ \end{gathered} \] yes! correct!

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1ok now we have to find the common denominator

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2right! we have to find the common denominator between the two denominators: \[\Large \begin{gathered} 3\left( {x  1} \right) \hfill \\ \left( {x  1} \right)\left( {x  2} \right) \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, we have to take all uncommon and common factors with the highest exponent

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1ok so how do we do that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is simple, x1 is a common factor, whereas 3 and x2 are uncommon factor, so the common denominator is: 3*(x1)*(x2)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and hence, we can write this: \[\Large \begin{gathered} \frac{{x + 1}}{{3x  3}}  \frac{4}{{{x^2}  3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x  1} \right)}}  \frac{4}{{\left( {x  1} \right)\left( {x  2} \right)}} = \hfill \\ \hfill \\ = \frac{{...}}{{3\left( {x  1} \right)\left( {x  2} \right)}} \hfill \\ \end{gathered} \]

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1so what is the common denominator

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is: \[\Large {3\left( {x  1} \right)\left( {x  2} \right)}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now we have to divide the common denominator by the first numerator, namely what is: \[\Large \frac{{3\left( {x  1} \right)\left( {x  2} \right)}}{{3\left( {x  1} \right)}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you can cancel out common factors

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440013189699:dw

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1why are you cancelling

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2because one factor is at denominator, and the same factor is at denominator

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please wait a moment

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the result is x2, so the next step is: \[\Large \begin{gathered} \frac{{x + 1}}{{3x  3}}  \frac{4}{{{x^2}  3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x  1} \right)}}  \frac{4}{{\left( {x  1} \right)\left( {x  2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x  2} \right)  4 \cdot ....}}{{3\left( {x  1} \right)\left( {x  2} \right)}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now we have to divide the common denominator by the second denominator, namely: \[\Large \frac{{3\left( {x  1} \right)\left( {x  2} \right)}}{{\left( {x  1} \right)\left( {x  2} \right)}} = ...\]

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1i dont think this is right

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! it is right :)

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1you have to do the common denominator for each part of the fraction

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the common denominator is only one

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1so for the first part wouldnt it be \[\frac{ x+1(x2) }{ 3(x1)(x2)}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, we have this: \[\Large \frac{{3\left( {x  1} \right)\left( {x  2} \right)}}{{\left( {x  1} \right)\left( {x  2} \right)}} = 3\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we can write this: \[\Large \begin{gathered} \frac{{x + 1}}{{3x  3}}  \frac{4}{{{x^2}  3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x  1} \right)}}  \frac{4}{{\left( {x  1} \right)\left( {x  2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x  2} \right)  4 \cdot 3}}{{3\left( {x  1} \right)\left( {x  2} \right)}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to simplify the numerator: \[\Large \begin{gathered} \frac{{x + 1}}{{3x  3}}  \frac{4}{{{x^2}  3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x  1} \right)}}  \frac{4}{{\left( {x  1} \right)\left( {x  2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x  2} \right)  4 \cdot 3}}{{3\left( {x  1} \right)\left( {x  2} \right)}} = \hfill \\ \hfill \\ = \frac{{{x^2}  x  2  12}}{{3\left( {x  1} \right)\left( {x  2} \right)}} = \frac{{{x^2}  x  14}}{{3\left( {x  1} \right)\left( {x  2} \right)}} \hfill \\ \end{gathered} \]

18jonea
 one year ago
Best ResponseYou've already chosen the best response.1and then that would be the final answer?
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